∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own
unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that satisfies G
in F.
G is false means there is a sequence of inference steps that satisfies G
in F.
*Thus the above G simply does not exist in F*
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own
unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that satisfies
G in F.
G is false means there is a sequence of inference steps that satisfies
G in F.
*Thus the above G simply does not exist in F*
So?
Since Godel's G isn't of that form, but only can be used to derive a
statment IN META-F that says that G is not provable in F, your argument
says nothing about Godel's G.
Also, you don't understand what those terms mean, because G being true doesn't mean there is no sequence of inference steps that satisfies G in
F, but there is no FINITE sequence of inference steps that satisfies G
in F.
Provable requires a FINITE sequence of steps
Being True just requires a sequence of connective steps, which can be infinite.
THus, a statement that is True but Unprovable says that there exist a sequence of connective steps, but that sequence is infinite in length,
so not usable as a proof.
You are just showing that you don't understand the meaning of the terms
you are using.
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own
unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that
satisfies G in F.
G is false means there is a sequence of inference steps that
satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link provability to
truth in the conventional model theory way.
I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.
Since Godel's G isn't of that form, but only can be used to derive a
statment IN META-F that says that G is not provable in F, your
argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐). https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
Also, you don't understand what those terms mean, because G being true
doesn't mean there is no sequence of inference steps that satisfies G
in F, but there is no FINITE sequence of inference steps that
satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can infer that G
is utterly unsatisfiable even for infinite sequences in this more
powerful F.
Provable requires a FINITE sequence of steps
Being True just requires a sequence of connective steps, which can be
infinite.
THus, a statement that is True but Unprovable says that there exist a
sequence of connective steps, but that sequence is infinite in length,
so not usable as a proof.
You are just showing that you don't understand the meaning of the
terms you are using.
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own
unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that
satisfies G in F.
G is false means there is a sequence of inference steps that
satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. You seem
to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you can't
actually understand even the outline of Godel's proof, so you take
pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own
unprovablity, as a simplification describing a statment DERIVED from G,
and that derivation happens in Meta-F, and is about what can be proven
in F.
Since Godel's G isn't of that form, but only can be used to derive a
statment IN META-F that says that G is not provable in F, your
argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G being
true doesn't mean there is no sequence of inference steps that
satisfies G in F, but there is no FINITE sequence of inference steps
that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can infer that G >> is utterly unsatisfiable even for infinite sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement.
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own
unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that
satisfies G in F.
G is false means there is a sequence of inference steps that
satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. You seem
to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you can't
actually understand even the outline of Godel's proof, so you take
pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own
unprovablity, as a simplification describing a statment DERIVED from
G, and that derivation happens in Meta-F, and is about what can be
proven in F.
Since Godel's G isn't of that form, but only can be used to derive a
statment IN META-F that says that G is not provable in F, your
argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G being
true doesn't mean there is no sequence of inference steps that
satisfies G in F, but there is no FINITE sequence of inference steps
that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can infer that G >>> is utterly unsatisfiable even for infinite sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own unprovability in F
A proof is any sequence of steps that shows that its conclusion is a necessary consequence of its premises.\
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own unprovability in F
If G is true then there is no sequence of inference steps that satisfies
G in F making G untrue.
If G is false then there is a sequence of inference steps that satisfies
G in F making G true.
Because the RHS of ↔ contradicts the LHS there is no such G in F.
Thus the above G simply does not exist in F.
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own
unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that
satisfies G in F.
G is false means there is a sequence of inference steps that
satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link provability to >>>> truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. You
seem to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you can't
actually understand even the outline of Godel's proof, so you take
pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own
unprovablity, as a simplification describing a statment DERIVED from
G, and that derivation happens in Meta-F, and is about what can be
proven in F.
Since Godel's G isn't of that form, but only can be used to derive
a statment IN META-F that says that G is not provable in F, your
argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom >>>> I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G being
true doesn't mean there is no sequence of inference steps that
satisfies G in F, but there is no FINITE sequence of inference
steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
that G
is utterly unsatisfiable even for infinite sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
A proof is any sequence of steps that shows that its conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given statement
is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
If G is true then there is no sequence of inference steps that
satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that
satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which forces G
to be true, thus this is a contradiction.
Because the RHS of ↔ contradicts the LHS there is no such G in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't finite, G can
be True but not Provable.
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own
unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that
satisfies G in F.
G is false means there is a sequence of inference steps that
satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link provability to >>>>> truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. You
seem to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you can't
actually understand even the outline of Godel's proof, so you take
pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own
unprovablity, as a simplification describing a statment DERIVED from
G, and that derivation happens in Meta-F, and is about what can be
proven in F.
Since Godel's G isn't of that form, but only can be used to derive >>>>>> a statment IN META-F that says that G is not provable in F, your
argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom >>>>> I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G being
true doesn't mean there is no sequence of inference steps that
satisfies G in F, but there is no FINITE sequence of inference
steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
that G
is utterly unsatisfiable even for infinite sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
A proof is any sequence of steps that shows that its conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given
statement is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
If G is true then there is no sequence of inference steps that
satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that
satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which forces
G to be true, thus this is a contradiction.
Because the RHS of ↔ contradicts the LHS there is no such G in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't finite, G
can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which negates the
RHS making it say (G ⊢ F) which makes G true in F.
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own
unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that
satisfies G in F.
G is false means there is a sequence of inference steps that
satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link provability to >>>> truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. You
seem to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you can't
actually understand even the outline of Godel's proof, so you take
pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own
unprovablity, as a simplification describing a statment DERIVED from
G, and that derivation happens in Meta-F, and is about what can be
proven in F.
Since Godel's G isn't of that form, but only can be used to derive
a statment IN META-F that says that G is not provable in F, your
argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom >>>> I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G being
true doesn't mean there is no sequence of inference steps that
satisfies G in F, but there is no FINITE sequence of inference
steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
that G
is utterly unsatisfiable even for infinite sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
A proof is any sequence of steps that shows that its conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given statement
is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
If G is true then there is no sequence of inference steps that
satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that
satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which forces G
to be true, thus this is a contradiction.
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own >>>>>>>> unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that
satisfies G in F.
G is false means there is a sequence of inference steps that
satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link
provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. You
seem to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you can't
actually understand even the outline of Godel's proof, so you take
pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own
unprovablity, as a simplification describing a statment DERIVED
from G, and that derivation happens in Meta-F, and is about what
can be proven in F.
Since Godel's G isn't of that form, but only can be used to
derive a statment IN META-F that says that G is not provable in
F, your argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom >>>>>> I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G being >>>>>>> true doesn't mean there is no sequence of inference steps that
satisfies G in F, but there is no FINITE sequence of inference
steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F that >>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can infer >>>>>> that G
is utterly unsatisfiable even for infinite sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
A proof is any sequence of steps that shows that its conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given
statement is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
If G is true then there is no sequence of inference steps that
satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that
satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which
forces G to be true, thus this is a contradiction.
Because the RHS of ↔ contradicts the LHS there is no such G in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't finite, G
can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which negates the
RHS making it say (G ⊢ F) which makes G true in F.
Right, G can't be false, but it can be True.
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its own
unprovability in F
*If we assume that there is such a G in F that means that*
G is true means there is no sequence of inference steps that
satisfies G in F.
G is false means there is a sequence of inference steps that
satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link provability to >>>>> truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. You
seem to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you can't
actually understand even the outline of Godel's proof, so you take
pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own
unprovablity, as a simplification describing a statment DERIVED from
G, and that derivation happens in Meta-F, and is about what can be
proven in F.
Since Godel's G isn't of that form, but only can be used to derive >>>>>> a statment IN META-F that says that G is not provable in F, your
argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom >>>>> I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G being
true doesn't mean there is no sequence of inference steps that
satisfies G in F, but there is no FINITE sequence of inference
steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can infer
that G
is utterly unsatisfiable even for infinite sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
A proof is any sequence of steps that shows that its conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given
statement is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its own
unprovability in F
If G is true then there is no sequence of inference steps that
satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that
satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which forces
G to be true, thus this is a contradiction.
When you agree with me on this very important point how does it make
sense to denigrate me further down?
Proving that there is at least one case where ↔ is not satisfied proves that there is no such G in F that satisfies ↔ in both cases, thus no
such G exists in F. Thus Gödel’s "incompleteness" theorem is transformed into Olcott's can't prove a contradiction theorem.
*It is like you are saying that I am a liar because I tell the truth*
I am never a Liar because Truth is the most important thing in my life
much more important than love.
Truth is the anchor of the basis of mathematically optimizing existence
thus creating paradise on Earth for all.
*It is like you are saying that I am a liar because I tell the truth*
I am never a Liar because Truth is the most important thing in my life
much more important than love.
Truth is the anchor of the basis of mathematically optimizing existence
thus creating paradise on Earth for all.
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its >>>>>>>>>> own unprovability in F
*If we assume that there is such a G in F that means that* >>>>>>>>>> G is true means there is no sequence of inference steps that >>>>>>>>>> satisfies G in F.
G is false means there is a sequence of inference steps that >>>>>>>>>> satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link
provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. You >>>>>>> seem to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F}
using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you
can't actually understand even the outline of Godel's proof, so
you take pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own
unprovablity, as a simplification describing a statment DERIVED
from G, and that derivation happens in Meta-F, and is about what >>>>>>> can be proven in F.
Since Godel's G isn't of that form, but only can be used to
derive a statment IN META-F that says that G is not provable in >>>>>>>>> F, your argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom >>>>>>>> I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G
being true doesn't mean there is no sequence of inference steps >>>>>>>>> that satisfies G in F, but there is no FINITE sequence of
inference steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F that >>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F can
infer that G
is utterly unsatisfiable even for infinite sequences in this more >>>>>>>> powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its
own unprovability in F
A proof is any sequence of steps that shows that its conclusion is a >>>>>> necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given
statement is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its
own unprovability in F
If G is true then there is no sequence of inference steps that
satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that
satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which
forces G to be true, thus this is a contradiction.
Because the RHS of ↔ contradicts the LHS there is no such G in F. >>>>>> Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't finite, G
can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which negates
the RHS making it say (G ⊢ F) which makes G true in F.
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is Unprovable.
A G that is ALWAYS True and ALWAYS unprovable statisfies that relationship.
"A if and only if B" doesn't requre that the case that neither A and B
being true exists.
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its >>>>>>>>>>> own unprovability in F
*If we assume that there is such a G in F that means that* >>>>>>>>>>> G is true means there is no sequence of inference steps that >>>>>>>>>>> satisfies G in F.
G is false means there is a sequence of inference steps that >>>>>>>>>>> satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link
provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference.
You seem to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F} >>>>>>>>> using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you
can't actually understand even the outline of Godel's proof, so >>>>>>>> you take pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own >>>>>>>> unprovablity, as a simplification describing a statment DERIVED >>>>>>>> from G, and that derivation happens in Meta-F, and is about what >>>>>>>> can be proven in F.
Since Godel's G isn't of that form, but only can be used to >>>>>>>>>> derive a statment IN META-F that says that G is not provable >>>>>>>>>> in F, your argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G >>>>>>>>>> being true doesn't mean there is no sequence of inference
steps that satisfies G in F, but there is no FINITE sequence >>>>>>>>>> of inference steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F >>>>>>>>> that
satisfies the RHS of ↔ contradicts the LHS a powerful F can >>>>>>>>> infer that G
is utterly unsatisfiable even for infinite sequences in this more >>>>>>>>> powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its >>>>>>> own unprovability in F
A proof is any sequence of steps that shows that its conclusion is a >>>>>>> necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given
statement is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its >>>>>>> own unprovability in F
If G is true then there is no sequence of inference steps that
satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE
sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that
satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which
forces G to be true, thus this is a contradiction.
Because the RHS of ↔ contradicts the LHS there is no such G in F. >>>>>>> Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't finite,
G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which negates >>>>> the RHS making it say (G ⊢ F) which makes G true in F.
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable. https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if G is unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if G is provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false.
A G that is ALWAYS True and ALWAYS unprovable statisfies that
relationship.
"A if and only if B" doesn't requre that the case that neither A and B
being true exists.
On 4/22/23 11:39 AM, olcott wrote:
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its >>>>>>>>>>>> own unprovability in F
*If we assume that there is such a G in F that means that* >>>>>>>>>>>> G is true means there is no sequence of inference steps that >>>>>>>>>>>> satisfies G in F.
G is false means there is a sequence of inference steps that >>>>>>>>>>>> satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link
provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. >>>>>>>>> You seem to confuse Truth with Knowledge.
I finally approximated {G asserts its own unprovability in F} >>>>>>>>>> using conventional math symbols in their conventional way.
Except that isn't what G is, you only think that because you >>>>>>>>> can't actually understand even the outline of Godel's proof, so >>>>>>>>> you take pieces out of context.
G never asserts its own unprovability.
The statement that we now have a statement that asserts its own >>>>>>>>> unprovablity, as a simplification describing a statment DERIVED >>>>>>>>> from G, and that derivation happens in Meta-F, and is about
what can be proven in F.
Since Godel's G isn't of that form, but only can be used to >>>>>>>>>>> derive a statment IN META-F that says that G is not provable >>>>>>>>>>> in F, your argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G >>>>>>>>>>> being true doesn't mean there is no sequence of inference >>>>>>>>>>> steps that satisfies G in F, but there is no FINITE sequence >>>>>>>>>>> of inference steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in F >>>>>>>>>> that
satisfies the RHS of ↔ contradicts the LHS a powerful F can >>>>>>>>>> infer that G
is utterly unsatisfiable even for infinite sequences in this more >>>>>>>>>> powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement. >>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its >>>>>>>> own unprovability in F
A proof is any sequence of steps that shows that its conclusion >>>>>>>> is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given
statement is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to its >>>>>>>> own unprovability in F
If G is true then there is no sequence of inference steps that >>>>>>>> satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an INFINITE >>>>>>> sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that >>>>>>>> satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which
forces G to be true, thus this is a contradiction.
Because the RHS of ↔ contradicts the LHS there is no such G in F. >>>>>>>> Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't finite, >>>>>>> G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which negates >>>>>> the RHS making it say (G ⊢ F) which makes G true in F.
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is
Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false.
But if neither row values can ACTUALLY EXIST, then the equality is true.
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to >>>>>>>>>>>>>> its own unprovability in F
*If we assume that there is such a G in F that means that* >>>>>>>>>>>>>> G is true means there is no sequence of inference steps >>>>>>>>>>>>>> that satisfies G in F.
G is false means there is a sequence of inference steps >>>>>>>>>>>>>> that satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link >>>>>>>>>>>> provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. >>>>>>>>>>> You seem to confuse Truth with Knowledge.
Except that isn't what G is, you only think that because you >>>>>>>>>>> can't actually understand even the outline of Godel's proof, >>>>>>>>>>> so you take pieces out of context.
I finally approximated {G asserts its own unprovability in F} >>>>>>>>>>>> using conventional math symbols in their conventional way. >>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts its >>>>>>>>>>> own unprovablity, as a simplification describing a statment >>>>>>>>>>> DERIVED from G, and that derivation happens in Meta-F, and is >>>>>>>>>>> about what can be proven in F.
Since Godel's G isn't of that form, but only can be used to >>>>>>>>>>>>> derive a statment IN META-F that says that G is not
provable in F, your argument says nothing about Godel's G. >>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G >>>>>>>>>>>>> being true doesn't mean there is no sequence of inference >>>>>>>>>>>>> steps that satisfies G in F, but there is no FINITE
sequence of inference steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in >>>>>>>>>>>> F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can >>>>>>>>>>>> infer that G
is utterly unsatisfiable even for infinite sequences in this >>>>>>>>>>>> more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement. >>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to >>>>>>>>>> its own unprovability in F
A proof is any sequence of steps that shows that its
conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given >>>>>>>>> statement is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to >>>>>>>>>> its own unprovability in F
If G is true then there is no sequence of inference steps that >>>>>>>>>> satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an
INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that >>>>>>>>>> satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which >>>>>>>>> forces G to be true, thus this is a contradiction.
Because the RHS of ↔ contradicts the LHS there is no such G in F. >>>>>>>>>> Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't
finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which
negates the RHS making it say (G ⊢ F) which makes G true in F. >>>>>>>>
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is
Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false.
But if neither row values can ACTUALLY EXIST, then the equality is true. >>>
(for whatever reason) then ↔ is unsatisfied and no such G exists in F.
You don't need to have all the rows with true being possible, you need
all the rows that are possible to be True.
More of your stupidity.
You just understand logic backwards.
On 4/22/2023 11:12 AM, Richard Damon wrote:
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to its >>>>>>>>>>>>> own unprovability in F
*If we assume that there is such a G in F that means that* >>>>>>>>>>>>> G is true means there is no sequence of inference steps >>>>>>>>>>>>> that satisfies G in F.
G is false means there is a sequence of inference steps >>>>>>>>>>>>> that satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link
provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the difference. >>>>>>>>>> You seem to confuse Truth with Knowledge.
Except that isn't what G is, you only think that because you >>>>>>>>>> can't actually understand even the outline of Godel's proof, >>>>>>>>>> so you take pieces out of context.
I finally approximated {G asserts its own unprovability in F} >>>>>>>>>>> using conventional math symbols in their conventional way. >>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts its >>>>>>>>>> own unprovablity, as a simplification describing a statment >>>>>>>>>> DERIVED from G, and that derivation happens in Meta-F, and is >>>>>>>>>> about what can be proven in F.
Since Godel's G isn't of that form, but only can be used to >>>>>>>>>>>> derive a statment IN META-F that says that G is not provable >>>>>>>>>>>> in F, your argument says nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because G >>>>>>>>>>>> being true doesn't mean there is no sequence of inference >>>>>>>>>>>> steps that satisfies G in F, but there is no FINITE sequence >>>>>>>>>>>> of inference steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence in >>>>>>>>>>> F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can >>>>>>>>>>> infer that G
is utterly unsatisfiable even for infinite sequences in this >>>>>>>>>>> more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement. >>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to >>>>>>>>> its own unprovability in F
A proof is any sequence of steps that shows that its conclusion >>>>>>>>> is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given
statement is a necessary consequence of the defined system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to >>>>>>>>> its own unprovability in F
If G is true then there is no sequence of inference steps that >>>>>>>>> satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an
INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps that >>>>>>>>> satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, which >>>>>>>> forces G to be true, thus this is a contradiction.
Because the RHS of ↔ contradicts the LHS there is no such G in F. >>>>>>>>> Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't
finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which
negates the RHS making it say (G ⊢ F) which makes G true in F. >>>>>>>
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is
Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false.
But if neither row values can ACTUALLY EXIST, then the equality is true.
(for whatever reason) then ↔ is unsatisfied and no such G exists in F.
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to >>>>>>>>>>>>>>>> its own unprovability in F
*If we assume that there is such a G in F that means that* >>>>>>>>>>>>>>>> G is true means there is no sequence of inference steps >>>>>>>>>>>>>>>> that satisfies G in F.
G is false means there is a sequence of inference steps >>>>>>>>>>>>>>>> that satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link >>>>>>>>>>>>>> provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the
difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>
Except that isn't what G is, you only think that because >>>>>>>>>>>>> you can't actually understand even the outline of Godel's >>>>>>>>>>>>> proof, so you take pieces out of context.
I finally approximated {G asserts its own unprovability in F} >>>>>>>>>>>>>> using conventional math symbols in their conventional way. >>>>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts its >>>>>>>>>>>>> own unprovablity, as a simplification describing a statment >>>>>>>>>>>>> DERIVED from G, and that derivation happens in Meta-F, and >>>>>>>>>>>>> is about what can be proven in F.
Since Godel's G isn't of that form, but only can be used >>>>>>>>>>>>>>> to derive a statment IN META-F that says that G is not >>>>>>>>>>>>>>> provable in F, your argument says nothing about Godel's G. >>>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because >>>>>>>>>>>>>>> G being true doesn't mean there is no sequence of >>>>>>>>>>>>>>> inference steps that satisfies G in F, but there is no >>>>>>>>>>>>>>> FINITE sequence of inference steps that satisfies G in F. >>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence >>>>>>>>>>>>>> in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F >>>>>>>>>>>>>> can infer that G
is utterly unsatisfiable even for infinite sequences in >>>>>>>>>>>>>> this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement. >>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to >>>>>>>>>>>> its own unprovability in F
A proof is any sequence of steps that shows that its
conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given >>>>>>>>>>> statement is a necessary consequence of the defined system. >>>>>>>>>>>
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when >>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to >>>>>>>>>>>> its own unprovability in F
If G is true then there is no sequence of inference steps >>>>>>>>>>>> that satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an >>>>>>>>>>> INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps >>>>>>>>>>>> that satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, >>>>>>>>>>> which forces G to be true, thus this is a contradiction. >>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS there is no such G >>>>>>>>>>>> in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't >>>>>>>>>>> finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which >>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F. >>>>>>>>>>
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is
Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if G is >>>>>> unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if G is >>>>>> provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false.
But if neither row values can ACTUALLY EXIST, then the equality is
true.
(for whatever reason) then ↔ is unsatisfied and no such G exists in F. >>>>
You don't need to have all the rows with true being possible, you
need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and the RHS
must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY values
that happen is G is ALWAYS true, an Unprovable is always true.
Thus the equivalence is always true.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to >>>>>>>>>>>>>>> its own unprovability in F
*If we assume that there is such a G in F that means that* >>>>>>>>>>>>>>> G is true means there is no sequence of inference steps >>>>>>>>>>>>>>> that satisfies G in F.
G is false means there is a sequence of inference steps >>>>>>>>>>>>>>> that satisfies G in F.
*Thus the above G simply does not exist in F*
So?
I finally learned enough model theory to correctly link >>>>>>>>>>>>> provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the
difference. You seem to confuse Truth with Knowledge.
Except that isn't what G is, you only think that because you >>>>>>>>>>>> can't actually understand even the outline of Godel's proof, >>>>>>>>>>>> so you take pieces out of context.
I finally approximated {G asserts its own unprovability in F} >>>>>>>>>>>>> using conventional math symbols in their conventional way. >>>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts its >>>>>>>>>>>> own unprovablity, as a simplification describing a statment >>>>>>>>>>>> DERIVED from G, and that derivation happens in Meta-F, and >>>>>>>>>>>> is about what can be proven in F.
Since Godel's G isn't of that form, but only can be used >>>>>>>>>>>>>> to derive a statment IN META-F that says that G is not >>>>>>>>>>>>>> provable in F, your argument says nothing about Godel's G. >>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to
Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, because >>>>>>>>>>>>>> G being true doesn't mean there is no sequence of
inference steps that satisfies G in F, but there is no >>>>>>>>>>>>>> FINITE sequence of inference steps that satisfies G in F. >>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence >>>>>>>>>>>>> in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F can >>>>>>>>>>>>> infer that G
is utterly unsatisfiable even for infinite sequences in >>>>>>>>>>>>> this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement. >>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to >>>>>>>>>>> its own unprovability in F
A proof is any sequence of steps that shows that its
conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a given >>>>>>>>>> statement is a necessary consequence of the defined system. >>>>>>>>>>
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent to >>>>>>>>>>> its own unprovability in F
If G is true then there is no sequence of inference steps >>>>>>>>>>> that satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an
INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps >>>>>>>>>>> that satisfies G in F making G true.
If G is false, then there is a finite sequence proving G,
which forces G to be true, thus this is a contradiction.
Because the RHS of ↔ contradicts the LHS there is no such G >>>>>>>>>>> in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't
finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which >>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F. >>>>>>>>>
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is
Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if G is >>>>> unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if G is >>>>> provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false.
But if neither row values can ACTUALLY EXIST, then the equality is
true.
(for whatever reason) then ↔ is unsatisfied and no such G exists in F. >>>
You don't need to have all the rows with true being possible, you need
all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and the RHS must always have the same truth value or ↔ is not true.
More of your stupidity.
You just understand logic backwards.
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent to >>>>>>>>>>>>>>>>> its own unprovability in F
*If we assume that there is such a G in F that means that* >>>>>>>>>>>>>>>>> G is true means there is no sequence of inference steps >>>>>>>>>>>>>>>>> that satisfies G in F.
G is false means there is a sequence of inference steps >>>>>>>>>>>>>>>>> that satisfies G in F.
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly link >>>>>>>>>>>>>>> provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the
difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>
Except that isn't what G is, you only think that because >>>>>>>>>>>>>> you can't actually understand even the outline of Godel's >>>>>>>>>>>>>> proof, so you take pieces out of context.
I finally approximated {G asserts its own unprovability >>>>>>>>>>>>>>> in F}
using conventional math symbols in their conventional way. >>>>>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts >>>>>>>>>>>>>> its own unprovablity, as a simplification describing a >>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in >>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
Since Godel's G isn't of that form, but only can be used >>>>>>>>>>>>>>>> to derive a statment IN META-F that says that G is not >>>>>>>>>>>>>>>> provable in F, your argument says nothing about Godel's G. >>>>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence >>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is >>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G >>>>>>>>>>>>>>>> in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite sequence >>>>>>>>>>>>>>> in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F >>>>>>>>>>>>>>> can infer that G
is utterly unsatisfiable even for infinite sequences in >>>>>>>>>>>>>>> this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement. >>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>> to its own unprovability in F
A proof is any sequence of steps that shows that its >>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a >>>>>>>>>>>> given statement is a necessary consequence of the defined >>>>>>>>>>>> system.
"Proof" doesn't have a "Premise", it has a system.
The statement may have conditions in it restricting when >>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>> to its own unprovability in F
If G is true then there is no sequence of inference steps >>>>>>>>>>>>> that satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an >>>>>>>>>>>> INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps >>>>>>>>>>>>> that satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, >>>>>>>>>>>> which forces G to be true, thus this is a contradiction. >>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS there is no such G >>>>>>>>>>>>> in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't >>>>>>>>>>>> finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which >>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F. >>>>>>>>>>>
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is
Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if G is >>>>>>> unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if >>>>>>> G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false.
But if neither row values can ACTUALLY EXIST, then the equality is >>>>>> true.
(for whatever reason) then ↔ is unsatisfied and no such G exists in F. >>>>>
You don't need to have all the rows with true being possible, you
need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and the RHS >>> must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY
values that happen is G is ALWAYS true, an Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS
We must assume that the RHS is false and see how that effects the LHS ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS) False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Right, and for that statement, the actual G found in F, the ONLYI don't think that is the way that it works.
values that happen is G is ALWAYS true, an Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS
We must assume that the RHS is false and see how that effects the LHS ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS) False(RHS) → True(LHS) refutes (RHS ↔ LHS)
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent >>>>>>>>>>>>>>>>>> to its own unprovability in F
*If we assume that there is such a G in F that means >>>>>>>>>>>>>>>>>> that*
G is true means there is no sequence of inference >>>>>>>>>>>>>>>>>> steps that satisfies G in F.
G is false means there is a sequence of inference >>>>>>>>>>>>>>>>>> steps that satisfies G in F.
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly link >>>>>>>>>>>>>>>> provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>
Except that isn't what G is, you only think that because >>>>>>>>>>>>>>> you can't actually understand even the outline of Godel's >>>>>>>>>>>>>>> proof, so you take pieces out of context.
I finally approximated {G asserts its own unprovability >>>>>>>>>>>>>>>> in F}
using conventional math symbols in their conventional way. >>>>>>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts >>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a >>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in >>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
Since Godel's G isn't of that form, but only can be >>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G is >>>>>>>>>>>>>>>>> not provable in F, your argument says nothing about >>>>>>>>>>>>>>>>> Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence >>>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is >>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G >>>>>>>>>>>>>>>>> in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F >>>>>>>>>>>>>>>> can infer that G
is utterly unsatisfiable even for infinite sequences in >>>>>>>>>>>>>>>> this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement. >>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>>> to its own unprovability in F
A proof is any sequence of steps that shows that its >>>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a >>>>>>>>>>>>> given statement is a necessary consequence of the defined >>>>>>>>>>>>> system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>
The statement may have conditions in it restricting when >>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>>> to its own unprovability in F
If G is true then there is no sequence of inference steps >>>>>>>>>>>>>> that satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an >>>>>>>>>>>>> INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps >>>>>>>>>>>>>> that satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, >>>>>>>>>>>>> which forces G to be true, thus this is a contradiction. >>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS there is no such >>>>>>>>>>>>>> G in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't >>>>>>>>>>>>> finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which >>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F. >>>>>>>>>>>>
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is >>>>>>>>> Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if >>>>>>>> G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if >>>>>>>> G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false.
But if neither row values can ACTUALLY EXIST, then the equality
is true.
(for whatever reason) then ↔ is unsatisfied and no such G exists >>>>>> in F.
You don't need to have all the rows with true being possible, you
need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and the RHS >>>> must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY
values that happen is G is ALWAYS true, an Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS
We must assume that the RHS is false and see how that effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent >>>>>>>>>>>>>>>>>> to its own unprovability in F
*If we assume that there is such a G in F that means >>>>>>>>>>>>>>>>>> that*
G is true means there is no sequence of inference >>>>>>>>>>>>>>>>>> steps that satisfies G in F.
G is false means there is a sequence of inference >>>>>>>>>>>>>>>>>> steps that satisfies G in F.
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly link >>>>>>>>>>>>>>>> provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>
Except that isn't what G is, you only think that because >>>>>>>>>>>>>>> you can't actually understand even the outline of Godel's >>>>>>>>>>>>>>> proof, so you take pieces out of context.
I finally approximated {G asserts its own unprovability >>>>>>>>>>>>>>>> in F}
using conventional math symbols in their conventional way. >>>>>>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts >>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a >>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in >>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
Since Godel's G isn't of that form, but only can be >>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G is >>>>>>>>>>>>>>>>> not provable in F, your argument says nothing about >>>>>>>>>>>>>>>>> Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence >>>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is >>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G >>>>>>>>>>>>>>>>> in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F >>>>>>>>>>>>>>>> can infer that G
is utterly unsatisfiable even for infinite sequences in >>>>>>>>>>>>>>>> this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do arguement. >>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>>> to its own unprovability in F
A proof is any sequence of steps that shows that its >>>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a >>>>>>>>>>>>> given statement is a necessary consequence of the defined >>>>>>>>>>>>> system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>
The statement may have conditions in it restricting when >>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>>> to its own unprovability in F
If G is true then there is no sequence of inference steps >>>>>>>>>>>>>> that satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an >>>>>>>>>>>>> INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps >>>>>>>>>>>>>> that satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, >>>>>>>>>>>>> which forces G to be true, thus this is a contradiction. >>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS there is no such >>>>>>>>>>>>>> G in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't >>>>>>>>>>>>> finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which >>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F. >>>>>>>>>>>>
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F.
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is >>>>>>>>> Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if >>>>>>>> G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if >>>>>>>> G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false.
But if neither row values can ACTUALLY EXIST, then the equality
is true.
(for whatever reason) then ↔ is unsatisfied and no such G exists >>>>>> in F.
You don't need to have all the rows with true being possible, you
need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and the RHS >>>> must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY
values that happen is G is ALWAYS true, an Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS
We must assume that the RHS is false and see how that effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q >>>>>>> (for whatever reason) then ↔ is unsatisfied and no such G exists >>>>>>> in F.
On 4/22/2023 9:57 AM, Richard Damon wrote:But if neither row values can ACTUALLY EXIST, then the equality >>>>>>>> is true.
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent >>>>>>>>>>>>>>>>>>> to its own unprovability in F
*If we assume that there is such a G in F that means >>>>>>>>>>>>>>>>>>> that*
G is true means there is no sequence of inference >>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
G is false means there is a sequence of inference >>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly link >>>>>>>>>>>>>>>>> provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>>
Except that isn't what G is, you only think that because >>>>>>>>>>>>>>>> you can't actually understand even the outline of >>>>>>>>>>>>>>>> Godel's proof, so you take pieces out of context. >>>>>>>>>>>>>>>>
I finally approximated {G asserts its own unprovability >>>>>>>>>>>>>>>>> in F}
using conventional math symbols in their conventional way. >>>>>>>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts >>>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a >>>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in >>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
Since Godel's G isn't of that form, but only can be >>>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G >>>>>>>>>>>>>>>>>> is not provable in F, your argument says nothing about >>>>>>>>>>>>>>>>>> Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence >>>>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is >>>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G >>>>>>>>>>>>>>>>>> in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F >>>>>>>>>>>>>>>>> can infer that G
is utterly unsatisfiable even for infinite sequences in >>>>>>>>>>>>>>>>> this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do >>>>>>>>>>>>>>>> arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>>>> to its own unprovability in F
A proof is any sequence of steps that shows that its >>>>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a >>>>>>>>>>>>>> given statement is a necessary consequence of the defined >>>>>>>>>>>>>> system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>
The statement may have conditions in it restricting when >>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>>>> to its own unprovability in F
If G is true then there is no sequence of inference steps >>>>>>>>>>>>>>> that satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an >>>>>>>>>>>>>> INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps >>>>>>>>>>>>>>> that satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, >>>>>>>>>>>>>> which forces G to be true, thus this is a contradiction. >>>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS there is no such >>>>>>>>>>>>>>> G in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't >>>>>>>>>>>>>> finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which >>>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F. >>>>>>>>>>>>>
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is >>>>>>>>>> Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if >>>>>>>>> G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only >>>>>>>>> if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false. >>>>>>>>
You don't need to have all the rows with true being possible, you
need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and the >>>>> RHS
must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY
values that happen is G is ALWAYS true, an Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS
We must assume that the RHS is false and see how that effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
I tried and in the first search all of the articles seemed to dodge
rather than address this point. I have always understood ↔ to mean that
the LHS and the RHS must always have the same Boolean value.
What sources do you have that directly contradict this?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Logical implication
p---q---p ⇒ q
T---T-----T // (p ⇒ q) ∧ (q ⇒ p) thus (p ↔ q)
T---F-----F // ¬(p ⇒ q) ∧ (q ⇒ p) thus ¬(p ↔ q)
F---T-----T // (p ⇒ q) ∧ ¬(q ⇒ p) thus ¬(p ↔ q)
F---F-----T // (p ⇒ q) ∧ (q ⇒ p) thus (p ↔ q)
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q >>>>>>>> (for whatever reason) then ↔ is unsatisfied and no such G exists >>>>>>>> in F.
On 4/22/2023 9:57 AM, Richard Damon wrote:But if neither row values can ACTUALLY EXIST, then the equality >>>>>>>>> is true.
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent >>>>>>>>>>>>>>>>>>>> to its own unprovability in F
*If we assume that there is such a G in F that means >>>>>>>>>>>>>>>>>>>> that*
G is true means there is no sequence of inference >>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
G is false means there is a sequence of inference >>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly >>>>>>>>>>>>>>>>>> link provability to
truth in the conventional model theory way. >>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own
unprovability in F}
using conventional math symbols in their conventional >>>>>>>>>>>>>>>>>> way.
Except that isn't what G is, you only think that >>>>>>>>>>>>>>>>> because you can't actually understand even the outline >>>>>>>>>>>>>>>>> of Godel's proof, so you take pieces out of context. >>>>>>>>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts >>>>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a >>>>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in >>>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F. >>>>>>>>>>>>>>>>>
Since Godel's G isn't of that form, but only can be >>>>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G >>>>>>>>>>>>>>>>>>> is not provable in F, your argument says nothing >>>>>>>>>>>>>>>>>>> about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no >>>>>>>>>>>>>>>>>>> sequence of inference steps that satisfies G in F, >>>>>>>>>>>>>>>>>>> but there is no FINITE sequence of inference steps >>>>>>>>>>>>>>>>>>> that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful >>>>>>>>>>>>>>>>>> F can infer that G
is utterly unsatisfiable even for infinite sequences >>>>>>>>>>>>>>>>>> in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do >>>>>>>>>>>>>>>>> arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>> equivalent to its own unprovability in F
A proof is any sequence of steps that shows that its >>>>>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a >>>>>>>>>>>>>>> given statement is a necessary consequence of the defined >>>>>>>>>>>>>>> system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>
The statement may have conditions in it restricting when >>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>> equivalent to its own unprovability in F
If G is true then there is no sequence of inference >>>>>>>>>>>>>>>> steps that satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an >>>>>>>>>>>>>>> INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference >>>>>>>>>>>>>>>> steps that satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, >>>>>>>>>>>>>>> which forces G to be true, thus this is a contradiction. >>>>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS there is no >>>>>>>>>>>>>>>> such G in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't >>>>>>>>>>>>>>> finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which >>>>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true >>>>>>>>>>>>>> in F.
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is >>>>>>>>>>> Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>
Row(1) There exists a G in F such that G is true if and only >>>>>>>>>> if G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only >>>>>>>>>> if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false. >>>>>>>>>
You don't need to have all the rows with true being possible, you >>>>>>> need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and
the RHS
must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY
values that happen is G is ALWAYS true, an Unprovable is always true. >>>>>
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS
We must assume that the RHS is false and see how that effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Here is a much clearer and conventional way of showing that
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this case
cause a problem.
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q >>>>>>> (for whatever reason) then ↔ is unsatisfied and no such G exists >>>>>>> in F.
On 4/22/2023 9:57 AM, Richard Damon wrote:But if neither row values can ACTUALLY EXIST, then the equality >>>>>>>> is true.
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote:
On 4/21/23 8:33 PM, olcott wrote:
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically equivalent >>>>>>>>>>>>>>>>>>> to its own unprovability in F
*If we assume that there is such a G in F that means >>>>>>>>>>>>>>>>>>> that*
G is true means there is no sequence of inference >>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
G is false means there is a sequence of inference >>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly link >>>>>>>>>>>>>>>>> provability to
truth in the conventional model theory way.
Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>>
Except that isn't what G is, you only think that because >>>>>>>>>>>>>>>> you can't actually understand even the outline of >>>>>>>>>>>>>>>> Godel's proof, so you take pieces out of context. >>>>>>>>>>>>>>>>
I finally approximated {G asserts its own unprovability >>>>>>>>>>>>>>>>> in F}
using conventional math symbols in their conventional way. >>>>>>>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that asserts >>>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a >>>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in >>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
Since Godel's G isn't of that form, but only can be >>>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G >>>>>>>>>>>>>>>>>> is not provable in F, your argument says nothing about >>>>>>>>>>>>>>>>>> Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence >>>>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is >>>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G >>>>>>>>>>>>>>>>>> in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful F >>>>>>>>>>>>>>>>> can infer that G
is utterly unsatisfiable even for infinite sequences in >>>>>>>>>>>>>>>>> this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do >>>>>>>>>>>>>>>> arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>>>> to its own unprovability in F
A proof is any sequence of steps that shows that its >>>>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a >>>>>>>>>>>>>> given statement is a necessary consequence of the defined >>>>>>>>>>>>>> system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>
The statement may have conditions in it restricting when >>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically equivalent >>>>>>>>>>>>>>> to its own unprovability in F
If G is true then there is no sequence of inference steps >>>>>>>>>>>>>>> that satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an >>>>>>>>>>>>>> INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference steps >>>>>>>>>>>>>>> that satisfies G in F making G true.
If G is false, then there is a finite sequence proving G, >>>>>>>>>>>>>> which forces G to be true, thus this is a contradiction. >>>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS there is no such >>>>>>>>>>>>>>> G in F.
Thus the above G simply does not exist in F.
Nope, because we can have an infinite sequence that isn't >>>>>>>>>>>>>> finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false which >>>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F. >>>>>>>>>>>>>
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G is >>>>>>>>>> Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if and only if >>>>>>>>> G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only >>>>>>>>> if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false. >>>>>>>>
You don't need to have all the rows with true being possible, you
need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and the >>>>> RHS
must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY
values that happen is G is ALWAYS true, an Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS
We must assume that the RHS is false and see how that effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Here is a much clearer and conventional way of showing that
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q >>>>>>>>> (for whatever reason) then ↔ is unsatisfied and no such G
On 4/22/2023 9:57 AM, Richard Damon wrote:But if neither row values can ACTUALLY EXIST, then the
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote:
On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F that >>>>>>>>>>>>>>>>>>>>> means that*
G is true means there is no sequence of inference >>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
G is false means there is a sequence of inference >>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly >>>>>>>>>>>>>>>>>>> link provability to
truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their conventional >>>>>>>>>>>>>>>>>>> way.
Except that isn't what G is, you only think that >>>>>>>>>>>>>>>>>> because you can't actually understand even the outline >>>>>>>>>>>>>>>>>> of Godel's proof, so you take pieces out of context. >>>>>>>>>>>>>>>>>>
G never asserts its own unprovability.
The statement that we now have a statement that >>>>>>>>>>>>>>>>>> asserts its own unprovablity, as a simplification >>>>>>>>>>>>>>>>>> describing a statment DERIVED from G, and that >>>>>>>>>>>>>>>>>> derivation happens in Meta-F, and is about what can be >>>>>>>>>>>>>>>>>> proven in F.
Since Godel's G isn't of that form, but only can be >>>>>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G >>>>>>>>>>>>>>>>>>>> is not provable in F, your argument says nothing >>>>>>>>>>>>>>>>>>>> about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no >>>>>>>>>>>>>>>>>>>> sequence of inference steps that satisfies G in F, >>>>>>>>>>>>>>>>>>>> but there is no FINITE sequence of inference steps >>>>>>>>>>>>>>>>>>>> that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a powerful >>>>>>>>>>>>>>>>>>> F can infer that G
is utterly unsatisfiable even for infinite sequences >>>>>>>>>>>>>>>>>>> in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do >>>>>>>>>>>>>>>>>> arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>> equivalent to its own unprovability in F
A proof is any sequence of steps that shows that its >>>>>>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a >>>>>>>>>>>>>>>> given statement is a necessary consequence of the >>>>>>>>>>>>>>>> defined system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>>
The statement may have conditions in it restricting when >>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>> equivalent to its own unprovability in F
If G is true then there is no sequence of inference >>>>>>>>>>>>>>>>> steps that satisfies G in F making G untrue.
no FINITE sequence, making G UNPROVABLE, and there IS an >>>>>>>>>>>>>>>> INFINITE sequence making it TRUE.
This is possible.
If G is false then there is a sequence of inference >>>>>>>>>>>>>>>>> steps that satisfies G in F making G true.
If G is false, then there is a finite sequence proving >>>>>>>>>>>>>>>> G, which forces G to be true, thus this is a contradiction. >>>>>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS there is no >>>>>>>>>>>>>>>>> such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence that >>>>>>>>>>>>>>>> isn't finite, G can be True but not Provable.
If G is false and ↔ is true this makes the RHS false >>>>>>>>>>>>>>> which negates the RHS making it say (G ⊢ F) which makes G >>>>>>>>>>>>>>> true in F.
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G >>>>>>>>>>>> is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>
Row(1) There exists a G in F such that G is true if and only >>>>>>>>>>> if G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only >>>>>>>>>>> if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false. >>>>>>>>>>
equality is true.
exists in F.
You don't need to have all the rows with true being possible,
you need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and >>>>>>> the RHS
must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY
values that happen is G is ALWAYS true, an Unprovable is always true. >>>>>>
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS >>>>> We must assume that the RHS is false and see how that effects the LHS >>>>> ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Here is a much clearer and conventional way of showing that
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this case
cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of logic the
above expression is simply false. There is no G that is logically
equivalent to its own unprovability in F.
On 4/22/23 3:34 PM, olcott wrote:
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p and q >>>>>>>>>> (for whatever reason) then ↔ is unsatisfied and no such G >>>>>>>>>> exists in F.
On 4/22/2023 9:57 AM, Richard Damon wrote:But if neither row values can ACTUALLY EXIST, then the
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F that >>>>>>>>>>>>>>>>>>>>>> means that*
G is true means there is no sequence of inference >>>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
G is false means there is a sequence of inference >>>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly >>>>>>>>>>>>>>>>>>>> link provability to
truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think that >>>>>>>>>>>>>>>>>>> because you can't actually understand even the >>>>>>>>>>>>>>>>>>> outline of Godel's proof, so you take pieces out of >>>>>>>>>>>>>>>>>>> context.
G never asserts its own unprovability.
The statement that we now have a statement that >>>>>>>>>>>>>>>>>>> asserts its own unprovablity, as a simplification >>>>>>>>>>>>>>>>>>> describing a statment DERIVED from G, and that >>>>>>>>>>>>>>>>>>> derivation happens in Meta-F, and is about what can >>>>>>>>>>>>>>>>>>> be proven in F.
Since Godel's G isn't of that form, but only can be >>>>>>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that >>>>>>>>>>>>>>>>>>>>> G is not provable in F, your argument says nothing >>>>>>>>>>>>>>>>>>>>> about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no >>>>>>>>>>>>>>>>>>>>> sequence of inference steps that satisfies G in F, >>>>>>>>>>>>>>>>>>>>> but there is no FINITE sequence of inference steps >>>>>>>>>>>>>>>>>>>>> that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a >>>>>>>>>>>>>>>>>>>> powerful F can infer that G
is utterly unsatisfiable even for infinite sequences >>>>>>>>>>>>>>>>>>>> in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do >>>>>>>>>>>>>>>>>>> arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F
A proof is any sequence of steps that shows that its >>>>>>>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that a >>>>>>>>>>>>>>>>> given statement is a necessary consequence of the >>>>>>>>>>>>>>>>> defined system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>>>
The statement may have conditions in it restricting when >>>>>>>>>>>>>>>>>
no FINITE sequence, making G UNPROVABLE, and there IS >>>>>>>>>>>>>>>>> an INFINITE sequence making it TRUE.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F
If G is true then there is no sequence of inference >>>>>>>>>>>>>>>>>> steps that satisfies G in F making G untrue. >>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of inference >>>>>>>>>>>>>>>>>> steps that satisfies G in F making G true.
If G is false, then there is a finite sequence proving >>>>>>>>>>>>>>>>> G, which forces G to be true, thus this is a >>>>>>>>>>>>>>>>> contradiction.
Because the RHS of ↔ contradicts the LHS there is no >>>>>>>>>>>>>>>>>> such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence that >>>>>>>>>>>>>>>>> isn't finite, G can be True but not Provable. >>>>>>>>>>>>>>>>>
If G is false and ↔ is true this makes the RHS false >>>>>>>>>>>>>>>> which negates the RHS making it say (G ⊢ F) which makes >>>>>>>>>>>>>>>> G true in F.
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G >>>>>>>>>>>>> is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>>
Row(1) There exists a G in F such that G is true if and only >>>>>>>>>>>> if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>
Row(4) There exists a G in F such that G is false if and >>>>>>>>>>>> only if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false. >>>>>>>>>>>
equality is true.
You don't need to have all the rows with true being possible, >>>>>>>>> you need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and >>>>>>>> the RHS
must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY >>>>>>> values that happen is G is ALWAYS true, an Unprovable is always
true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS >>>>>> We must assume that the RHS is false and see how that effects the LHS >>>>>> ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Here is a much clearer and conventional way of showing that
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this case
cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of logic the
above expression is simply false. There is no G that is logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a number g such
that g statisifes a particular Primative Recursive Relationship (built
in Meta-F, but using only operations defined in F).
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not falsely place
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p >>>>>>>>>>> and q
On 4/22/2023 9:57 AM, Richard Damon wrote:But if neither row values can ACTUALLY EXIST, then the >>>>>>>>>>>> equality is true.
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote:
On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>>>
There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F that >>>>>>>>>>>>>>>>>>>>>>> means that*
G is true means there is no sequence of inference >>>>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
G is false means there is a sequence of inference >>>>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly >>>>>>>>>>>>>>>>>>>>> link provability to
truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think that >>>>>>>>>>>>>>>>>>>> because you can't actually understand even the >>>>>>>>>>>>>>>>>>>> outline of Godel's proof, so you take pieces out of >>>>>>>>>>>>>>>>>>>> context.
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement that >>>>>>>>>>>>>>>>>>>> asserts its own unprovablity, as a simplification >>>>>>>>>>>>>>>>>>>> describing a statment DERIVED from G, and that >>>>>>>>>>>>>>>>>>>> derivation happens in Meta-F, and is about what can >>>>>>>>>>>>>>>>>>>> be proven in F.
Since Godel's G isn't of that form, but only can >>>>>>>>>>>>>>>>>>>>>> be used to derive a statment IN META-F that says >>>>>>>>>>>>>>>>>>>>>> that G is not provable in F, your argument says >>>>>>>>>>>>>>>>>>>>>> nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no >>>>>>>>>>>>>>>>>>>>>> sequence of inference steps that satisfies G in F, >>>>>>>>>>>>>>>>>>>>>> but there is no FINITE sequence of inference steps >>>>>>>>>>>>>>>>>>>>>> that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a >>>>>>>>>>>>>>>>>>>>> powerful F can infer that G
is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>> sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do >>>>>>>>>>>>>>>>>>>> arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that shows that its >>>>>>>>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\
Boy are you wrong.
A proof is a FINITE sequence of steps that shows that >>>>>>>>>>>>>>>>>> a given statement is a necessary consequence of the >>>>>>>>>>>>>>>>>> defined system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>>>>
The statement may have conditions in it restricting when >>>>>>>>>>>>>>>>>>
no FINITE sequence, making G UNPROVABLE, and there IS >>>>>>>>>>>>>>>>>> an INFINITE sequence making it TRUE.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of inference >>>>>>>>>>>>>>>>>>> steps that satisfies G in F making G untrue. >>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of inference >>>>>>>>>>>>>>>>>>> steps that satisfies G in F making G true. >>>>>>>>>>>>>>>>>>If G is false, then there is a finite sequence proving >>>>>>>>>>>>>>>>>> G, which forces G to be true, thus this is a >>>>>>>>>>>>>>>>>> contradiction.
Because the RHS of ↔ contradicts the LHS there is no >>>>>>>>>>>>>>>>>>> such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence that >>>>>>>>>>>>>>>>>> isn't finite, G can be True but not Provable. >>>>>>>>>>>>>>>>>>
If G is false and ↔ is true this makes the RHS false >>>>>>>>>>>>>>>>> which negates the RHS making it say (G ⊢ F) which makes >>>>>>>>>>>>>>>>> G true in F.
Right, G can't be false, but it can be True.
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if G >>>>>>>>>>>>>> is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>>>
Row(1) There exists a G in F such that G is true if and >>>>>>>>>>>>> only if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>>
Row(4) There exists a G in F such that G is false if and >>>>>>>>>>>>> only if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false. >>>>>>>>>>>>
(for whatever reason) then ↔ is unsatisfied and no such G >>>>>>>>>>> exists in F.
You don't need to have all the rows with true being possible, >>>>>>>>>> you need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS and >>>>>>>>> the RHS
must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the ONLY >>>>>>>> values that happen is G is ALWAYS true, an Unprovable is always >>>>>>>> true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the LHS >>>>>>> We must assume that the RHS is false and see how that effects the >>>>>>> LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Here is a much clearer and conventional way of showing that
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this case
cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of logic the
above expression is simply false. There is no G that is logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a number g such
that g statisifes a particular Primative Recursive Relationship (built
in Meta-F, but using only operations defined in F).
the blame on F.
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not falsely place
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p >>>>>>>>>>>> and q
On 4/22/2023 9:57 AM, Richard Damon wrote:But if neither row values can ACTUALLY EXIST, then the >>>>>>>>>>>>> equality is true.
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:Boy are you wrong.
On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>>>>
There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F that >>>>>>>>>>>>>>>>>>>>>>>> means that*
G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to correctly >>>>>>>>>>>>>>>>>>>>>> link provability to
truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think that >>>>>>>>>>>>>>>>>>>>> because you can't actually understand even the >>>>>>>>>>>>>>>>>>>>> outline of Godel's proof, so you take pieces out of >>>>>>>>>>>>>>>>>>>>> context.
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement that >>>>>>>>>>>>>>>>>>>>> asserts its own unprovablity, as a simplification >>>>>>>>>>>>>>>>>>>>> describing a statment DERIVED from G, and that >>>>>>>>>>>>>>>>>>>>> derivation happens in Meta-F, and is about what can >>>>>>>>>>>>>>>>>>>>> be proven in F.
Since Godel's G isn't of that form, but only can >>>>>>>>>>>>>>>>>>>>>>> be used to derive a statment IN META-F that says >>>>>>>>>>>>>>>>>>>>>>> that G is not provable in F, your argument says >>>>>>>>>>>>>>>>>>>>>>> nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐).
https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms mean, >>>>>>>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no >>>>>>>>>>>>>>>>>>>>>>> sequence of inference steps that satisfies G in >>>>>>>>>>>>>>>>>>>>>>> F, but there is no FINITE sequence of inference >>>>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a >>>>>>>>>>>>>>>>>>>>>> powerful F can infer that G
is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>>> sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do >>>>>>>>>>>>>>>>>>>>> arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that shows that its >>>>>>>>>>>>>>>>>>>> conclusion is a
necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that shows that >>>>>>>>>>>>>>>>>>> a given statement is a necessary consequence of the >>>>>>>>>>>>>>>>>>> defined system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>>>>>
The statement may have conditions in it restricting when >>>>>>>>>>>>>>>>>>>
no FINITE sequence, making G UNPROVABLE, and there IS >>>>>>>>>>>>>>>>>>> an INFINITE sequence making it TRUE.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of inference >>>>>>>>>>>>>>>>>>>> steps that satisfies G in F making G untrue. >>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of inference >>>>>>>>>>>>>>>>>>>> steps that satisfies G in F making G true. >>>>>>>>>>>>>>>>>>>If G is false, then there is a finite sequence >>>>>>>>>>>>>>>>>>> proving G, which forces G to be true, thus this is a >>>>>>>>>>>>>>>>>>> contradiction.
Because the RHS of ↔ contradicts the LHS there is no >>>>>>>>>>>>>>>>>>>> such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence that >>>>>>>>>>>>>>>>>>> isn't finite, G can be True but not Provable. >>>>>>>>>>>>>>>>>>>
If G is false and ↔ is true this makes the RHS false >>>>>>>>>>>>>>>>>> which negates the RHS making it say (G ⊢ F) which >>>>>>>>>>>>>>>>>> makes G true in F.
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if >>>>>>>>>>>>>>> G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>>>>
Row(1) There exists a G in F such that G is true if and >>>>>>>>>>>>>> only if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>>>
Row(4) There exists a G in F such that G is false if and >>>>>>>>>>>>>> only if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false. >>>>>>>>>>>>>
(for whatever reason) then ↔ is unsatisfied and no such G >>>>>>>>>>>> exists in F.
You don't need to have all the rows with true being possible, >>>>>>>>>>> you need all the rows that are possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS >>>>>>>>>> and the RHS
must always have the same truth value or ↔ is not true.
Right, and for that statement, the actual G found in F, the
ONLY values that happen is G is ALWAYS true, an Unprovable is >>>>>>>>> always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects the >>>>>>>> LHS
We must assume that the RHS is false and see how that effects
the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Here is a much clearer and conventional way of showing that
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this case
cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of logic the >>>> above expression is simply false. There is no G that is logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a number g
such that g statisifes a particular Primative Recursive Relationship
(built in Meta-F, but using only operations defined in F).
the blame on F.
Yes, but can you PROVE your statement? If not, you are just making unsubstantiated false claims, just like DT.
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not falsely place
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for p >>>>>>>>>>>>> and q
On 4/22/2023 9:57 AM, Richard Damon wrote:But if neither row values can ACTUALLY EXIST, then the >>>>>>>>>>>>>> equality is true.
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote:
On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>Boy are you wrong.
Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>>>>>There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F that >>>>>>>>>>>>>>>>>>>>>>>>> means that*
G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>> correctly link provability to
truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think that >>>>>>>>>>>>>>>>>>>>>> because you can't actually understand even the >>>>>>>>>>>>>>>>>>>>>> outline of Godel's proof, so you take pieces out >>>>>>>>>>>>>>>>>>>>>> of context.
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement that >>>>>>>>>>>>>>>>>>>>>> asserts its own unprovablity, as a simplification >>>>>>>>>>>>>>>>>>>>>> describing a statment DERIVED from G, and that >>>>>>>>>>>>>>>>>>>>>> derivation happens in Meta-F, and is about what >>>>>>>>>>>>>>>>>>>>>> can be proven in F.
Since Godel's G isn't of that form, but only can >>>>>>>>>>>>>>>>>>>>>>>> be used to derive a statment IN META-F that says >>>>>>>>>>>>>>>>>>>>>>>> that G is not provable in F, your argument says >>>>>>>>>>>>>>>>>>>>>>>> nothing about Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
So?
Did you read that article?
Also, you don't understand what those terms >>>>>>>>>>>>>>>>>>>>>>>> mean, because G being true doesn't mean there is >>>>>>>>>>>>>>>>>>>>>>>> no sequence of inference steps that satisfies G >>>>>>>>>>>>>>>>>>>>>>>> in F, but there is no FINITE sequence of >>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a >>>>>>>>>>>>>>>>>>>>>>> powerful F can infer that G
is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>>>> sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do >>>>>>>>>>>>>>>>>>>>>> arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that shows that >>>>>>>>>>>>>>>>>>>>> its conclusion is a
necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that shows >>>>>>>>>>>>>>>>>>>> that a given statement is a necessary consequence of >>>>>>>>>>>>>>>>>>>> the defined system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>>>>>>
The statement may have conditions in it restricting >>>>>>>>>>>>>>>>>>>> when
no FINITE sequence, making G UNPROVABLE, and there >>>>>>>>>>>>>>>>>>>> IS an INFINITE sequence making it TRUE. >>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of inference >>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F making G untrue. >>>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of inference >>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F making G true. >>>>>>>>>>>>>>>>>>>>If G is false, then there is a finite sequence >>>>>>>>>>>>>>>>>>>> proving G, which forces G to be true, thus this is a >>>>>>>>>>>>>>>>>>>> contradiction.
Because the RHS of ↔ contradicts the LHS there is >>>>>>>>>>>>>>>>>>>>> no such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence that >>>>>>>>>>>>>>>>>>>> isn't finite, G can be True but not Provable. >>>>>>>>>>>>>>>>>>>>
If G is false and ↔ is true this makes the RHS false >>>>>>>>>>>>>>>>>>> which negates the RHS making it say (G ⊢ F) which >>>>>>>>>>>>>>>>>>> makes G true in F.
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only if >>>>>>>>>>>>>>>> G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>>>>>
Row(1) There exists a G in F such that G is true if and >>>>>>>>>>>>>>> only if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>>>>
Row(4) There exists a G in F such that G is false if and >>>>>>>>>>>>>>> only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then ↔ is false. >>>>>>>>>>>>>>
(for whatever reason) then ↔ is unsatisfied and no such G >>>>>>>>>>>>> exists in F.
You don't need to have all the rows with true being
possible, you need all the rows that are possible to be True. >>>>>>>>>>>>
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS >>>>>>>>>>> and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>
Right, and for that statement, the actual G found in F, the >>>>>>>>>> ONLY values that happen is G is ALWAYS true, an Unprovable is >>>>>>>>>> always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects >>>>>>>>> the LHS
We must assume that the RHS is false and see how that effects >>>>>>>>> the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔
LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Here is a much clearer and conventional way of showing that
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this
case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of logic the >>>>> above expression is simply false. There is no G that is logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a number g
such that g statisifes a particular Primative Recursive Relationship
(built in Meta-F, but using only operations defined in F).
the blame on F.
Yes, but can you PROVE your statement? If not, you are just making
unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of
understanding (an honest mistake not a lie) about how ↔ works.
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not falsely place >>>> the blame on F.
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for >>>>>>>>>>>>>> p and q
On 4/22/2023 9:57 AM, Richard Damon wrote:But if neither row values can ACTUALLY EXIST, then the >>>>>>>>>>>>>>> equality is true.
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>Boy are you wrong.
Doesn't seem so, you don't seem to understand the >>>>>>>>>>>>>>>>>>>>>>> difference. You seem to confuse Truth with >>>>>>>>>>>>>>>>>>>>>>> Knowledge.There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F that >>>>>>>>>>>>>>>>>>>>>>>>>> means that*
G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to
truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think that >>>>>>>>>>>>>>>>>>>>>>> because you can't actually understand even the >>>>>>>>>>>>>>>>>>>>>>> outline of Godel's proof, so you take pieces out >>>>>>>>>>>>>>>>>>>>>>> of context.
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement that >>>>>>>>>>>>>>>>>>>>>>> asserts its own unprovablity, as a simplification >>>>>>>>>>>>>>>>>>>>>>> describing a statment DERIVED from G, and that >>>>>>>>>>>>>>>>>>>>>>> derivation happens in Meta-F, and is about what >>>>>>>>>>>>>>>>>>>>>>> can be proven in F.
So?
Since Godel's G isn't of that form, but only >>>>>>>>>>>>>>>>>>>>>>>>> can be used to derive a statment IN META-F that >>>>>>>>>>>>>>>>>>>>>>>>> says that G is not provable in F, your argument >>>>>>>>>>>>>>>>>>>>>>>>> says nothing about Godel's G. >>>>>>>>>>>>>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>
Did you read that article?
Also, you don't understand what those terms >>>>>>>>>>>>>>>>>>>>>>>>> mean, because G being true doesn't mean there >>>>>>>>>>>>>>>>>>>>>>>>> is no sequence of inference steps that >>>>>>>>>>>>>>>>>>>>>>>>> satisfies G in F, but there is no FINITE >>>>>>>>>>>>>>>>>>>>>>>>> sequence of inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F))
Because we can see that every finite or infinite >>>>>>>>>>>>>>>>>>>>>>>> sequence in F that
satisfies the RHS of ↔ contradicts the LHS a >>>>>>>>>>>>>>>>>>>>>>>> powerful F can infer that G
is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>>>>> sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only do >>>>>>>>>>>>>>>>>>>>>>> arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that shows that >>>>>>>>>>>>>>>>>>>>>> its conclusion is a
necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that shows >>>>>>>>>>>>>>>>>>>>> that a given statement is a necessary consequence >>>>>>>>>>>>>>>>>>>>> of the defined system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>>>>>>>
The statement may have conditions in it restricting >>>>>>>>>>>>>>>>>>>>> when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G >>>>>>>>>>>>>>>>>>>>>> untrue.
no FINITE sequence, making G UNPROVABLE, and there >>>>>>>>>>>>>>>>>>>>> IS an INFINITE sequence making it TRUE. >>>>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G true. >>>>>>>>>>>>>>>>>>>>>If G is false, then there is a finite sequence >>>>>>>>>>>>>>>>>>>>> proving G, which forces G to be true, thus this is >>>>>>>>>>>>>>>>>>>>> a contradiction.
Because the RHS of ↔ contradicts the LHS there is >>>>>>>>>>>>>>>>>>>>>> no such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence that >>>>>>>>>>>>>>>>>>>>> isn't finite, G can be True but not Provable. >>>>>>>>>>>>>>>>>>>>>
If G is false and ↔ is true this makes the RHS false >>>>>>>>>>>>>>>>>>>> which negates the RHS making it say (G ⊢ F) which >>>>>>>>>>>>>>>>>>>> makes G true in F.
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean.
There exists a G in F such that G is true if and only >>>>>>>>>>>>>>>>> if G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>>>>>>
Row(1) There exists a G in F such that G is true if and >>>>>>>>>>>>>>>> only if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>>>>>
Row(4) There exists a G in F such that G is false if and >>>>>>>>>>>>>>>> only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then ↔ is false. >>>>>>>>>>>>>>>
(for whatever reason) then ↔ is unsatisfied and no such G >>>>>>>>>>>>>> exists in F.
You don't need to have all the rows with true being
possible, you need all the rows that are possible to be True. >>>>>>>>>>>>>
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS >>>>>>>>>>>> and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>>
Right, and for that statement, the actual G found in F, the >>>>>>>>>>> ONLY values that happen is G is ALWAYS true, an Unprovable is >>>>>>>>>>> always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects >>>>>>>>>> the LHS
We must assume that the RHS is false and see how that effects >>>>>>>>>> the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔
LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Here is a much clearer and conventional way of showing that
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this
case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of logic >>>>>> the
above expression is simply false. There is no G that is logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a number g
such that g statisifes a particular Primative Recursive
Relationship (built in Meta-F, but using only operations defined in
F).
Yes, but can you PROVE your statement? If not, you are just making
unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of
understanding (an honest mistake not a lie) about how ↔ works.
Nope, how did you prove that no such G exists? You claims that row 4
can't be satisfied? it doesn't need to ever be used.
On 4/22/2023 3:06 PM, Richard Damon wrote:
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not falsely
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value for >>>>>>>>>>>>>>> p and q
On 4/22/2023 9:57 AM, Richard Damon wrote:
On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>
Boy are you wrong.Doesn't seem so, you don't seem to understand >>>>>>>>>>>>>>>>>>>>>>>> the difference. You seem to confuse Truth with >>>>>>>>>>>>>>>>>>>>>>>> Knowledge.There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F >>>>>>>>>>>>>>>>>>>>>>>>>>> that means that*
G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to >>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think that >>>>>>>>>>>>>>>>>>>>>>>> because you can't actually understand even the >>>>>>>>>>>>>>>>>>>>>>>> outline of Godel's proof, so you take pieces out >>>>>>>>>>>>>>>>>>>>>>>> of context.
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement that >>>>>>>>>>>>>>>>>>>>>>>> asserts its own unprovablity, as a >>>>>>>>>>>>>>>>>>>>>>>> simplification describing a statment DERIVED >>>>>>>>>>>>>>>>>>>>>>>> from G, and that derivation happens in Meta-F, >>>>>>>>>>>>>>>>>>>>>>>> and is about what can be proven in F. >>>>>>>>>>>>>>>>>>>>>>>>
So?
Since Godel's G isn't of that form, but only >>>>>>>>>>>>>>>>>>>>>>>>>> can be used to derive a statment IN META-F >>>>>>>>>>>>>>>>>>>>>>>>>> that says that G is not provable in F, your >>>>>>>>>>>>>>>>>>>>>>>>>> argument says nothing about Godel's G. >>>>>>>>>>>>>>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>>
Did you read that article?
Also, you don't understand what those terms >>>>>>>>>>>>>>>>>>>>>>>>>> mean, because G being true doesn't mean there >>>>>>>>>>>>>>>>>>>>>>>>>> is no sequence of inference steps that >>>>>>>>>>>>>>>>>>>>>>>>>> satisfies G in F, but there is no FINITE >>>>>>>>>>>>>>>>>>>>>>>>>> sequence of inference steps that satisfies G >>>>>>>>>>>>>>>>>>>>>>>>>> in F.
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>
Because we can see that every finite or >>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that
satisfies the RHS of ↔ contradicts the LHS a >>>>>>>>>>>>>>>>>>>>>>>>> powerful F can infer that G
is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>>>>>> sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only >>>>>>>>>>>>>>>>>>>>>>>> do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that shows that >>>>>>>>>>>>>>>>>>>>>>> its conclusion is a
necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that shows >>>>>>>>>>>>>>>>>>>>>> that a given statement is a necessary consequence >>>>>>>>>>>>>>>>>>>>>> of the defined system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>>>>>>>>
The statement may have conditions in it >>>>>>>>>>>>>>>>>>>>>> restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G >>>>>>>>>>>>>>>>>>>>>>> untrue.
no FINITE sequence, making G UNPROVABLE, and there >>>>>>>>>>>>>>>>>>>>>> IS an INFINITE sequence making it TRUE. >>>>>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G true. >>>>>>>>>>>>>>>>>>>>>>If G is false, then there is a finite sequence >>>>>>>>>>>>>>>>>>>>>> proving G, which forces G to be true, thus this is >>>>>>>>>>>>>>>>>>>>>> a contradiction.
Because the RHS of ↔ contradicts the LHS there is >>>>>>>>>>>>>>>>>>>>>>> no such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence >>>>>>>>>>>>>>>>>>>>>> that isn't finite, G can be True but not Provable. >>>>>>>>>>>>>>>>>>>>>>
If G is false and ↔ is true this makes the RHS >>>>>>>>>>>>>>>>>>>>> false which negates the RHS making it say (G ⊢ F) >>>>>>>>>>>>>>>>>>>>> which makes G true in F.
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean. >>>>>>>>>>>>>>>>>>
There exists a G in F such that G is true if and only >>>>>>>>>>>>>>>>>> if G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>>>>>>>
Row(1) There exists a G in F such that G is true if and >>>>>>>>>>>>>>>>> only if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>>>>>>
Row(4) There exists a G in F such that G is false if >>>>>>>>>>>>>>>>> and only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then ↔ is >>>>>>>>>>>>>>>>> false.
But if neither row values can ACTUALLY EXIST, then the >>>>>>>>>>>>>>>> equality is true.
(for whatever reason) then ↔ is unsatisfied and no such G >>>>>>>>>>>>>>> exists in F.
You don't need to have all the rows with true being >>>>>>>>>>>>>> possible, you need all the rows that are possible to be True. >>>>>>>>>>>>>>
To the best of my knowledge
↔ is also known as logical equivalence meaning that the LHS >>>>>>>>>>>>> and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>>>
Right, and for that statement, the actual G found in F, the >>>>>>>>>>>> ONLY values that happen is G is ALWAYS true, an Unprovable >>>>>>>>>>>> is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects >>>>>>>>>>> the LHS
We must assume that the RHS is false and see how that effects >>>>>>>>>>> the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS
↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works?
p ↔ q would seem to mean ((p → q) ∧ (q → p))
Here is a much clearer and conventional way of showing that
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this >>>>>>>> case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of
logic the
above expression is simply false. There is no G that is logically >>>>>>> equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a number g
such that g statisifes a particular Primative Recursive
Relationship (built in Meta-F, but using only operations defined
in F).
place
the blame on F.
Yes, but can you PROVE your statement? If not, you are just making
unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of
understanding (an honest mistake not a lie) about how ↔ works.
Nope, how did you prove that no such G exists? You claims that row 4
can't be satisfied? it doesn't need to ever be used.
Try and prove that with a source, in the mean time I will tentatively
assume that you are wrong. I proved that I am correct with the above
truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p))
On 4/22/2023 3:14 PM, Richard Damon wrote:
On 4/22/23 4:10 PM, olcott wrote:I may have been mistaken when I thought that more than one row of the
On 4/22/2023 3:06 PM, Richard Damon wrote:
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not falsely >>>>>>> place
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:p ↔ q would seem to mean ((p → q) ∧ (q → p))
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>> On 4/22/23 11:39 AM, olcott wrote:So, you don't understand how truth tables work. >>>>>>>>>>>>>>>>
If either Row(1) or Row(4) cannot have the same value >>>>>>>>>>>>>>>>> for p and qOn 4/22/2023 9:57 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Boy are you wrong.Doesn't seem so, you don't seem to understand >>>>>>>>>>>>>>>>>>>>>>>>>> the difference. You seem to confuse Truth with >>>>>>>>>>>>>>>>>>>>>>>>>> Knowledge.There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>> that means that*
G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to >>>>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think >>>>>>>>>>>>>>>>>>>>>>>>>> that because you can't actually understand >>>>>>>>>>>>>>>>>>>>>>>>>> even the outline of Godel's proof, so you take >>>>>>>>>>>>>>>>>>>>>>>>>> pieces out of context.
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement >>>>>>>>>>>>>>>>>>>>>>>>>> that asserts its own unprovablity, as a >>>>>>>>>>>>>>>>>>>>>>>>>> simplification describing a statment DERIVED >>>>>>>>>>>>>>>>>>>>>>>>>> from G, and that derivation happens in Meta-F, >>>>>>>>>>>>>>>>>>>>>>>>>> and is about what can be proven in F. >>>>>>>>>>>>>>>>>>>>>>>>>>
So?
Since Godel's G isn't of that form, but only >>>>>>>>>>>>>>>>>>>>>>>>>>>> can be used to derive a statment IN META-F >>>>>>>>>>>>>>>>>>>>>>>>>>>> that says that G is not provable in F, your >>>>>>>>>>>>>>>>>>>>>>>>>>>> argument says nothing about Godel's G. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>>>>
Did you read that article? >>>>>>>>>>>>>>>>>>>>>>>>>>
Also, you don't understand what those terms >>>>>>>>>>>>>>>>>>>>>>>>>>>> mean, because G being true doesn't mean >>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no sequence of inference steps that >>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies G in F, but there is no FINITE >>>>>>>>>>>>>>>>>>>>>>>>>>>> sequence of inference steps that satisfies G >>>>>>>>>>>>>>>>>>>>>>>>>>>> in F.
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>
Because we can see that every finite or >>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that >>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a >>>>>>>>>>>>>>>>>>>>>>>>>>> powerful F can infer that G >>>>>>>>>>>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>>>>>>>> sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only >>>>>>>>>>>>>>>>>>>>>>>>>> do arguement.
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that shows >>>>>>>>>>>>>>>>>>>>>>>>> that its conclusion is a
necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that shows >>>>>>>>>>>>>>>>>>>>>>>> that a given statement is a necessary >>>>>>>>>>>>>>>>>>>>>>>> consequence of the defined system. >>>>>>>>>>>>>>>>>>>>>>>>
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>>>>>>>>>>
The statement may have conditions in it >>>>>>>>>>>>>>>>>>>>>>>> restricting when
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G >>>>>>>>>>>>>>>>>>>>>>>>> untrue.
no FINITE sequence, making G UNPROVABLE, and >>>>>>>>>>>>>>>>>>>>>>>> there IS an INFINITE sequence making it TRUE. >>>>>>>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G >>>>>>>>>>>>>>>>>>>>>>>>> true.
If G is false, then there is a finite sequence >>>>>>>>>>>>>>>>>>>>>>>> proving G, which forces G to be true, thus this >>>>>>>>>>>>>>>>>>>>>>>> is a contradiction.
Because the RHS of ↔ contradicts the LHS there >>>>>>>>>>>>>>>>>>>>>>>>> is no such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence >>>>>>>>>>>>>>>>>>>>>>>> that isn't finite, G can be True but not Provable. >>>>>>>>>>>>>>>>>>>>>>>>
If G is false and ↔ is true this makes the RHS >>>>>>>>>>>>>>>>>>>>>>> false which negates the RHS making it say (G ⊢ F) >>>>>>>>>>>>>>>>>>>>>>> which makes G true in F.
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean. >>>>>>>>>>>>>>>>>>>>
There exists a G in F such that G is true if and >>>>>>>>>>>>>>>>>>>> only if G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>>>>>>>>>
Row(1) There exists a G in F such that G is true if >>>>>>>>>>>>>>>>>>> and only if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>>>>>>>>
Row(4) There exists a G in F such that G is false if >>>>>>>>>>>>>>>>>>> and only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then ↔ is >>>>>>>>>>>>>>>>>>> false.
But if neither row values can ACTUALLY EXIST, then the >>>>>>>>>>>>>>>>>> equality is true.
(for whatever reason) then ↔ is unsatisfied and no such >>>>>>>>>>>>>>>>> G exists in F.
You don't need to have all the rows with true being >>>>>>>>>>>>>>>> possible, you need all the rows that are possible to be >>>>>>>>>>>>>>>> True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the >>>>>>>>>>>>>>> LHS and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>>>>>
Right, and for that statement, the actual G found in F, >>>>>>>>>>>>>> the ONLY values that happen is G is ALWAYS true, an >>>>>>>>>>>>>> Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that >>>>>>>>>>>>> effects the LHS
We must assume that the RHS is false and see how that >>>>>>>>>>>>> effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ >>>>>>>>>>>>> (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works? >>>>>>>>>>>
Here is a much clearer and conventional way of showing that >>>>>>>>>>>
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this >>>>>>>>>> case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of >>>>>>>>> logic the
above expression is simply false. There is no G that is logically >>>>>>>>> equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a number >>>>>>>> g such that g statisifes a particular Primative Recursive
Relationship (built in Meta-F, but using only operations defined >>>>>>>> in F).
the blame on F.
Yes, but can you PROVE your statement? If not, you are just making >>>>>> unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of
understanding (an honest mistake not a lie) about how ↔ works.
Nope, how did you prove that no such G exists? You claims that row 4
can't be satisfied? it doesn't need to ever be used.
Try and prove that with a source, in the mean time I will tentatively
assume that you are wrong. I proved that I am correct with the above
truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p))
WRONG, YOU are making the claim, so YOU need to prove it.
truth table needed to be satisfied. Furthermore in retrospect this looks
like a dumb mistake that I did not notice as a dumb mistake until I
looked at the truth table for ∧. So we are back to row one.
∃G ∈ F (G ↔ (F ⊬ G))
If the RHS is satisfied then this means that there are no inference
steps in F that derive G, thus G cannot be shown to be true in F.
On 4/22/23 4:10 PM, olcott wrote:
On 4/22/2023 3:06 PM, Richard Damon wrote:
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not falsely
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:p ↔ q would seem to mean ((p → q) ∧ (q → p))
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote:So, you don't understand how truth tables work.
On 4/22/23 11:39 AM, olcott wrote:If either Row(1) or Row(4) cannot have the same value >>>>>>>>>>>>>>>> for p and q
On 4/22/2023 9:57 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Boy are you wrong.Doesn't seem so, you don't seem to understand >>>>>>>>>>>>>>>>>>>>>>>>> the difference. You seem to confuse Truth with >>>>>>>>>>>>>>>>>>>>>>>>> Knowledge.There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>> that means that*
G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to >>>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think >>>>>>>>>>>>>>>>>>>>>>>>> that because you can't actually understand even >>>>>>>>>>>>>>>>>>>>>>>>> the outline of Godel's proof, so you take >>>>>>>>>>>>>>>>>>>>>>>>> pieces out of context.
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement that >>>>>>>>>>>>>>>>>>>>>>>>> asserts its own unprovablity, as a >>>>>>>>>>>>>>>>>>>>>>>>> simplification describing a statment DERIVED >>>>>>>>>>>>>>>>>>>>>>>>> from G, and that derivation happens in Meta-F, >>>>>>>>>>>>>>>>>>>>>>>>> and is about what can be proven in F. >>>>>>>>>>>>>>>>>>>>>>>>>
So?
Since Godel's G isn't of that form, but only >>>>>>>>>>>>>>>>>>>>>>>>>>> can be used to derive a statment IN META-F >>>>>>>>>>>>>>>>>>>>>>>>>>> that says that G is not provable in F, your >>>>>>>>>>>>>>>>>>>>>>>>>>> argument says nothing about Godel's G. >>>>>>>>>>>>>>>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is equivalent to >>>>>>>>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>>>
Did you read that article?
Also, you don't understand what those terms >>>>>>>>>>>>>>>>>>>>>>>>>>> mean, because G being true doesn't mean there >>>>>>>>>>>>>>>>>>>>>>>>>>> is no sequence of inference steps that >>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies G in F, but there is no FINITE >>>>>>>>>>>>>>>>>>>>>>>>>>> sequence of inference steps that satisfies G >>>>>>>>>>>>>>>>>>>>>>>>>>> in F.
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>
Because we can see that every finite or >>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that >>>>>>>>>>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a >>>>>>>>>>>>>>>>>>>>>>>>>> powerful F can infer that G >>>>>>>>>>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>>>>>>> sequences in this more
powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can only >>>>>>>>>>>>>>>>>>>>>>>>> do arguement.
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that shows that >>>>>>>>>>>>>>>>>>>>>>>> its conclusion is a
necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that shows >>>>>>>>>>>>>>>>>>>>>>> that a given statement is a necessary consequence >>>>>>>>>>>>>>>>>>>>>>> of the defined system.
"Proof" doesn't have a "Premise", it has a system. >>>>>>>>>>>>>>>>>>>>>>>
The statement may have conditions in it >>>>>>>>>>>>>>>>>>>>>>> restricting when
∃G ∈ F (G ↔ (G ⊬ F))
There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G >>>>>>>>>>>>>>>>>>>>>>>> untrue.
no FINITE sequence, making G UNPROVABLE, and >>>>>>>>>>>>>>>>>>>>>>> there IS an INFINITE sequence making it TRUE. >>>>>>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G >>>>>>>>>>>>>>>>>>>>>>>> true.
If G is false, then there is a finite sequence >>>>>>>>>>>>>>>>>>>>>>> proving G, which forces G to be true, thus this >>>>>>>>>>>>>>>>>>>>>>> is a contradiction.
Because the RHS of ↔ contradicts the LHS there >>>>>>>>>>>>>>>>>>>>>>>> is no such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence >>>>>>>>>>>>>>>>>>>>>>> that isn't finite, G can be True but not Provable. >>>>>>>>>>>>>>>>>>>>>>>
If G is false and ↔ is true this makes the RHS >>>>>>>>>>>>>>>>>>>>>> false which negates the RHS making it say (G ⊢ F) >>>>>>>>>>>>>>>>>>>>>> which makes G true in F.
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G exists in F. >>>>>>>>>>>>>>>>>>>>
Why do you say that?
I don't think you know what you terms mean. >>>>>>>>>>>>>>>>>>>
There exists a G in F such that G is true if and only >>>>>>>>>>>>>>>>>>> if G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>>>>>>>>
Row(1) There exists a G in F such that G is true if >>>>>>>>>>>>>>>>>> and only if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>>>>>>>
Row(4) There exists a G in F such that G is false if >>>>>>>>>>>>>>>>>> and only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then ↔ is >>>>>>>>>>>>>>>>>> false.
But if neither row values can ACTUALLY EXIST, then the >>>>>>>>>>>>>>>>> equality is true.
(for whatever reason) then ↔ is unsatisfied and no such >>>>>>>>>>>>>>>> G exists in F.
You don't need to have all the rows with true being >>>>>>>>>>>>>>> possible, you need all the rows that are possible to be >>>>>>>>>>>>>>> True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the >>>>>>>>>>>>>> LHS and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>>>>
Right, and for that statement, the actual G found in F, the >>>>>>>>>>>>> ONLY values that happen is G is ALWAYS true, an Unprovable >>>>>>>>>>>>> is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that effects >>>>>>>>>>>> the LHS
We must assume that the RHS is false and see how that
effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS
↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works? >>>>>>>>>>
Here is a much clearer and conventional way of showing that >>>>>>>>>>
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in this >>>>>>>>> case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of
logic the
above expression is simply false. There is no G that is logically >>>>>>>> equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a number g >>>>>>> such that g statisifes a particular Primative Recursive
Relationship (built in Meta-F, but using only operations defined >>>>>>> in F).
place
the blame on F.
Yes, but can you PROVE your statement? If not, you are just making
unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of
understanding (an honest mistake not a lie) about how ↔ works.
Nope, how did you prove that no such G exists? You claims that row 4
can't be satisfied? it doesn't need to ever be used.
Try and prove that with a source, in the mean time I will tentatively
assume that you are wrong. I proved that I am correct with the above
truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p))
WRONG, YOU are making the claim, so YOU need to prove it.
On 4/22/2023 3:43 PM, Richard Damon wrote:
On 4/22/23 4:39 PM, olcott wrote:
On 4/22/2023 3:14 PM, Richard Damon wrote:
On 4/22/23 4:10 PM, olcott wrote:I may have been mistaken when I thought that more than one row of the
On 4/22/2023 3:06 PM, Richard Damon wrote:
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:p ↔ q would seem to mean ((p → q) ∧ (q → p)) >>>>>>>>>>>>> Here is a much clearer and conventional way of showing that >>>>>>>>>>>>>
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>> On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 4/22/23 11:39 AM, olcott wrote:So, you don't understand how truth tables work. >>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) cannot have the same value >>>>>>>>>>>>>>>>>>> for p and qOn 4/22/2023 9:57 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Boy are you wrong.Doesn't seem so, you don't seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>> understand the difference. You seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>> confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>>>>>>>>>>>There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that means that* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F*
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think >>>>>>>>>>>>>>>>>>>>>>>>>>>> that because you can't actually understand >>>>>>>>>>>>>>>>>>>>>>>>>>>> even the outline of Godel's proof, so you >>>>>>>>>>>>>>>>>>>>>>>>>>>> take pieces out of context. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement >>>>>>>>>>>>>>>>>>>>>>>>>>>> that asserts its own unprovablity, as a >>>>>>>>>>>>>>>>>>>>>>>>>>>> simplification describing a statment DERIVED >>>>>>>>>>>>>>>>>>>>>>>>>>>> from G, and that derivation happens in >>>>>>>>>>>>>>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
Since Godel's G isn't of that form, but >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> only can be used to derive a statment IN >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> META-F that says that G is not provable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> F, your argument says nothing about >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to
Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Did you read that article? >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Also, you don't understand what those >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> terms mean, because G being true doesn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mean there is no sequence of inference >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F, but there is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because we can see that every finite or >>>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that >>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS >>>>>>>>>>>>>>>>>>>>>>>>>>>>> a powerful F can infer that G >>>>>>>>>>>>>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequences in this more >>>>>>>>>>>>>>>>>>>>>>>>>>>>> powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can >>>>>>>>>>>>>>>>>>>>>>>>>>>> only do arguement.
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own unprovability >>>>>>>>>>>>>>>>>>>>>>>>>>> in F
A proof is any sequence of steps that shows >>>>>>>>>>>>>>>>>>>>>>>>>>> that its conclusion is a >>>>>>>>>>>>>>>>>>>>>>>>>>> necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that >>>>>>>>>>>>>>>>>>>>>>>>>> shows that a given statement is a necessary >>>>>>>>>>>>>>>>>>>>>>>>>> consequence of the defined system. >>>>>>>>>>>>>>>>>>>>>>>>>>
"Proof" doesn't have a "Premise", it has a >>>>>>>>>>>>>>>>>>>>>>>>>> system.
The statement may have conditions in it >>>>>>>>>>>>>>>>>>>>>>>>>> restricting when
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own unprovability >>>>>>>>>>>>>>>>>>>>>>>>>>> in F
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making >>>>>>>>>>>>>>>>>>>>>>>>>>> G untrue.
no FINITE sequence, making G UNPROVABLE, and >>>>>>>>>>>>>>>>>>>>>>>>>> there IS an INFINITE sequence making it TRUE. >>>>>>>>>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making >>>>>>>>>>>>>>>>>>>>>>>>>>> G true.
If G is false, then there is a finite sequence >>>>>>>>>>>>>>>>>>>>>>>>>> proving G, which forces G to be true, thus >>>>>>>>>>>>>>>>>>>>>>>>>> this is a contradiction.
Because the RHS of ↔ contradicts the LHS >>>>>>>>>>>>>>>>>>>>>>>>>>> there is no such G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>> Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence >>>>>>>>>>>>>>>>>>>>>>>>>> that isn't finite, G can be True but not >>>>>>>>>>>>>>>>>>>>>>>>>> Provable.
If G is false and ↔ is true this makes the RHS >>>>>>>>>>>>>>>>>>>>>>>>> false which negates the RHS making it say (G ⊢ >>>>>>>>>>>>>>>>>>>>>>>>> F) which makes G true in F.
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G exists >>>>>>>>>>>>>>>>>>>>>>> in F.
Why do you say that?
I don't think you know what you terms mean. >>>>>>>>>>>>>>>>>>>>>>
There exists a G in F such that G is true if and >>>>>>>>>>>>>>>>>>>>>> only if G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true if >>>>>>>>>>>>>>>>>>>>> and only if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>>>>>>>>>>
Row(4) There exists a G in F such that G is false >>>>>>>>>>>>>>>>>>>>> if and only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then ↔ >>>>>>>>>>>>>>>>>>>>> is false.
But if neither row values can ACTUALLY EXIST, then >>>>>>>>>>>>>>>>>>>> the equality is true.
(for whatever reason) then ↔ is unsatisfied and no >>>>>>>>>>>>>>>>>>> such G exists in F.
You don't need to have all the rows with true being >>>>>>>>>>>>>>>>>> possible, you need all the rows that are possible to >>>>>>>>>>>>>>>>>> be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the >>>>>>>>>>>>>>>>> LHS and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>>>>>>>
Right, and for that statement, the actual G found in F, >>>>>>>>>>>>>>>> the ONLY values that happen is G is ALWAYS true, an >>>>>>>>>>>>>>>> Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that >>>>>>>>>>>>>>> effects the LHS
We must assume that the RHS is false and see how that >>>>>>>>>>>>>>> effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ >>>>>>>>>>>>>>> (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works? >>>>>>>>>>>>>
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in >>>>>>>>>>>> this case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of >>>>>>>>>>> logic the
above expression is simply false. There is no G that is
logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a
number g such that g statisifes a particular Primative
Recursive Relationship (built in Meta-F, but using only
operations defined in F).
falsely place
the blame on F.
Yes, but can you PROVE your statement? If not, you are just
making unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of
understanding (an honest mistake not a lie) about how ↔ works. >>>>>>>
Nope, how did you prove that no such G exists? You claims that row >>>>>> 4 can't be satisfied? it doesn't need to ever be used.
Try and prove that with a source, in the mean time I will tentatively >>>>> assume that you are wrong. I proved that I am correct with the above >>>>> truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p)) >>>>>
WRONG, YOU are making the claim, so YOU need to prove it.
truth table needed to be satisfied. Furthermore in retrospect this looks >>> like a dumb mistake that I did not notice as a dumb mistake until I
looked at the truth table for ∧. So we are back to row one.
∃G ∈ F (G ↔ (F ⊬ G))
If the RHS is satisfied then this means that there are no inference
steps in F that derive G, thus G cannot be shown to be true in F.
Nope, there is no FINITE series of infernece steps in F that derive G.
This G cannot be shown to be true in F.
There can be an INFINITE series of inference steps in F that derive G,
making it True but unprovable.
You already said that there cannot be infinite inference steps in F.
You are just continuing to show that you don't understand what "Proof"
means.
On 4/22/23 4:39 PM, olcott wrote:
On 4/22/2023 3:14 PM, Richard Damon wrote:
On 4/22/23 4:10 PM, olcott wrote:I may have been mistaken when I thought that more than one row of the
On 4/22/2023 3:06 PM, Richard Damon wrote:
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:p ↔ q would seem to mean ((p → q) ∧ (q → p))
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote:
On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 4/22/23 11:39 AM, olcott wrote:So, you don't understand how truth tables work. >>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) cannot have the same value >>>>>>>>>>>>>>>>>> for p and qOn 4/22/2023 9:57 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Boy are you wrong.Doesn't seem so, you don't seem to understand >>>>>>>>>>>>>>>>>>>>>>>>>>> the difference. You seem to confuse Truth >>>>>>>>>>>>>>>>>>>>>>>>>>> with Knowledge.There exists a G such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that means that*
G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist in F* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to >>>>>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F}
using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think >>>>>>>>>>>>>>>>>>>>>>>>>>> that because you can't actually understand >>>>>>>>>>>>>>>>>>>>>>>>>>> even the outline of Godel's proof, so you >>>>>>>>>>>>>>>>>>>>>>>>>>> take pieces out of context. >>>>>>>>>>>>>>>>>>>>>>>>>>>
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement >>>>>>>>>>>>>>>>>>>>>>>>>>> that asserts its own unprovablity, as a >>>>>>>>>>>>>>>>>>>>>>>>>>> simplification describing a statment DERIVED >>>>>>>>>>>>>>>>>>>>>>>>>>> from G, and that derivation happens in >>>>>>>>>>>>>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
Since Godel's G isn't of that form, but >>>>>>>>>>>>>>>>>>>>>>>>>>>>> only can be used to derive a statment IN >>>>>>>>>>>>>>>>>>>>>>>>>>>>> META-F that says that G is not provable in >>>>>>>>>>>>>>>>>>>>>>>>>>>>> F, your argument says nothing about Godel's G. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is >>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to
Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>>>>>
Did you read that article? >>>>>>>>>>>>>>>>>>>>>>>>>>>
Also, you don't understand what those terms >>>>>>>>>>>>>>>>>>>>>>>>>>>>> mean, because G being true doesn't mean >>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no sequence of inference steps >>>>>>>>>>>>>>>>>>>>>>>>>>>>> that satisfies G in F, but there is no >>>>>>>>>>>>>>>>>>>>>>>>>>>>> FINITE sequence of inference steps that >>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies G in F.
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because we can see that every finite or >>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that >>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a >>>>>>>>>>>>>>>>>>>>>>>>>>>> powerful F can infer that G >>>>>>>>>>>>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>>>>>>>>> sequences in this more >>>>>>>>>>>>>>>>>>>>>>>>>>>> powerful F.
Nope. Show the PROOF.
You don't know HOW to do a proof, you can >>>>>>>>>>>>>>>>>>>>>>>>>>> only do arguement.
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that shows >>>>>>>>>>>>>>>>>>>>>>>>>> that its conclusion is a
necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that >>>>>>>>>>>>>>>>>>>>>>>>> shows that a given statement is a necessary >>>>>>>>>>>>>>>>>>>>>>>>> consequence of the defined system. >>>>>>>>>>>>>>>>>>>>>>>>>
"Proof" doesn't have a "Premise", it has a >>>>>>>>>>>>>>>>>>>>>>>>> system.
The statement may have conditions in it >>>>>>>>>>>>>>>>>>>>>>>>> restricting when
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is logically >>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to its own unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G >>>>>>>>>>>>>>>>>>>>>>>>>> untrue.
no FINITE sequence, making G UNPROVABLE, and >>>>>>>>>>>>>>>>>>>>>>>>> there IS an INFINITE sequence making it TRUE. >>>>>>>>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making G >>>>>>>>>>>>>>>>>>>>>>>>>> true.
If G is false, then there is a finite sequence >>>>>>>>>>>>>>>>>>>>>>>>> proving G, which forces G to be true, thus this >>>>>>>>>>>>>>>>>>>>>>>>> is a contradiction.
Because the RHS of ↔ contradicts the LHS there >>>>>>>>>>>>>>>>>>>>>>>>>> is no such G in F.
Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite sequence >>>>>>>>>>>>>>>>>>>>>>>>> that isn't finite, G can be True but not Provable. >>>>>>>>>>>>>>>>>>>>>>>>>
If G is false and ↔ is true this makes the RHS >>>>>>>>>>>>>>>>>>>>>>>> false which negates the RHS making it say (G ⊢ >>>>>>>>>>>>>>>>>>>>>>>> F) which makes G true in F.
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G exists >>>>>>>>>>>>>>>>>>>>>> in F.
Why do you say that?
I don't think you know what you terms mean. >>>>>>>>>>>>>>>>>>>>>
There exists a G in F such that G is true if and >>>>>>>>>>>>>>>>>>>>> only if G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality >>>>>>>>>>>>>>>>>>>>
Row(1) There exists a G in F such that G is true if >>>>>>>>>>>>>>>>>>>> and only if G is
unprovable in F making G unsatisfied thus untrue in F. >>>>>>>>>>>>>>>>>>>>
Row(4) There exists a G in F such that G is false if >>>>>>>>>>>>>>>>>>>> and only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then ↔ is >>>>>>>>>>>>>>>>>>>> false.
But if neither row values can ACTUALLY EXIST, then >>>>>>>>>>>>>>>>>>> the equality is true.
(for whatever reason) then ↔ is unsatisfied and no >>>>>>>>>>>>>>>>>> such G exists in F.
You don't need to have all the rows with true being >>>>>>>>>>>>>>>>> possible, you need all the rows that are possible to be >>>>>>>>>>>>>>>>> True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that the >>>>>>>>>>>>>>>> LHS and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>>>>>>
Right, and for that statement, the actual G found in F, >>>>>>>>>>>>>>> the ONLY values that happen is G is ALWAYS true, an >>>>>>>>>>>>>>> Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that >>>>>>>>>>>>>> effects the LHS
We must assume that the RHS is false and see how that >>>>>>>>>>>>>> effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ >>>>>>>>>>>>>> (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS)
Nope, that isn't how it works.
Can you show me something that says that is how it works? >>>>>>>>>>>>
Here is a much clearer and conventional way of showing that >>>>>>>>>>>>
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in >>>>>>>>>>> this case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of >>>>>>>>>> logic the
above expression is simply false. There is no G that is logically >>>>>>>>>> equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a number >>>>>>>>> g such that g statisifes a particular Primative Recursive
Relationship (built in Meta-F, but using only operations
defined in F).
falsely place
the blame on F.
Yes, but can you PROVE your statement? If not, you are just
making unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of
understanding (an honest mistake not a lie) about how ↔ works.
Nope, how did you prove that no such G exists? You claims that row
4 can't be satisfied? it doesn't need to ever be used.
Try and prove that with a source, in the mean time I will tentatively
assume that you are wrong. I proved that I am correct with the above
truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p))
WRONG, YOU are making the claim, so YOU need to prove it.
truth table needed to be satisfied. Furthermore in retrospect this looks
like a dumb mistake that I did not notice as a dumb mistake until I
looked at the truth table for ∧. So we are back to row one.
∃G ∈ F (G ↔ (F ⊬ G))
If the RHS is satisfied then this means that there are no inference
steps in F that derive G, thus G cannot be shown to be true in F.
Nope, there is no FINITE series of infernece steps in F that derive G.
There can be an INFINITE series of inference steps in F that derive G,
making it True but unprovable.
You are just continuing to show that you don't understand what "Proof"
means.
On 4/22/23 4:47 PM, olcott wrote:
On 4/22/2023 3:43 PM, Richard Damon wrote:
On 4/22/23 4:39 PM, olcott wrote:
On 4/22/2023 3:14 PM, Richard Damon wrote:
On 4/22/23 4:10 PM, olcott wrote:I may have been mistaken when I thought that more than one row of the
On 4/22/2023 3:06 PM, Richard Damon wrote:
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:p ↔ q would seem to mean ((p → q) ∧ (q → p)) >>>>>>>>>>>>>> Here is a much clearer and conventional way of showing that >>>>>>>>>>>>>>
On 4/22/2023 11:56 AM, Richard Damon wrote:
On 4/22/23 12:45 PM, olcott wrote:I don't think that is the way that it works.
On 4/22/2023 11:36 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 4/22/23 11:39 AM, olcott wrote:So, you don't understand how truth tables work. >>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) cannot have the same >>>>>>>>>>>>>>>>>>>> value for p and qOn 4/22/2023 9:57 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Boy are you wrong.Doesn't seem so, you don't seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand the difference. You seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>There exists a G such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> F that means that* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F*
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F} >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> conventional way.
Except that isn't what G is, you only think >>>>>>>>>>>>>>>>>>>>>>>>>>>>> that because you can't actually understand >>>>>>>>>>>>>>>>>>>>>>>>>>>>> even the outline of Godel's proof, so you >>>>>>>>>>>>>>>>>>>>>>>>>>>>> take pieces out of context. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement >>>>>>>>>>>>>>>>>>>>>>>>>>>>> that asserts its own unprovablity, as a >>>>>>>>>>>>>>>>>>>>>>>>>>>>> simplification describing a statment >>>>>>>>>>>>>>>>>>>>>>>>>>>>> DERIVED from G, and that derivation happens >>>>>>>>>>>>>>>>>>>>>>>>>>>>> in Meta-F, and is about what can be proven >>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F.
So?
Since Godel's G isn't of that form, but >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> only can be used to derive a statment IN >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> META-F that says that G is not provable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F, your argument says nothing about >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to
Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Did you read that article? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Also, you don't understand what those >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> terms mean, because G being true doesn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mean there is no sequence of inference >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F, but there is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> no FINITE sequence of inference steps >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because we can see that every finite or >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a powerful F can infer that G >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequences in this more >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> powerful F.
Nope. Show the PROOF. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't know HOW to do a proof, you can >>>>>>>>>>>>>>>>>>>>>>>>>>>>> only do arguement.
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F
A proof is any sequence of steps that shows >>>>>>>>>>>>>>>>>>>>>>>>>>>> that its conclusion is a >>>>>>>>>>>>>>>>>>>>>>>>>>>> necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that >>>>>>>>>>>>>>>>>>>>>>>>>>> shows that a given statement is a necessary >>>>>>>>>>>>>>>>>>>>>>>>>>> consequence of the defined system. >>>>>>>>>>>>>>>>>>>>>>>>>>>
"Proof" doesn't have a "Premise", it has a >>>>>>>>>>>>>>>>>>>>>>>>>>> system.
The statement may have conditions in it >>>>>>>>>>>>>>>>>>>>>>>>>>> restricting when
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making >>>>>>>>>>>>>>>>>>>>>>>>>>>> G untrue.
no FINITE sequence, making G UNPROVABLE, and >>>>>>>>>>>>>>>>>>>>>>>>>>> there IS an INFINITE sequence making it TRUE. >>>>>>>>>>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F making >>>>>>>>>>>>>>>>>>>>>>>>>>>> G true.
If G is false, then there is a finite >>>>>>>>>>>>>>>>>>>>>>>>>>> sequence proving G, which forces G to be >>>>>>>>>>>>>>>>>>>>>>>>>>> true, thus this is a contradiction. >>>>>>>>>>>>>>>>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS >>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no such G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>> Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite >>>>>>>>>>>>>>>>>>>>>>>>>>> sequence that isn't finite, G can be True but >>>>>>>>>>>>>>>>>>>>>>>>>>> not Provable.
If G is false and ↔ is true this makes the RHS >>>>>>>>>>>>>>>>>>>>>>>>>> false which negates the RHS making it say (G ⊢ >>>>>>>>>>>>>>>>>>>>>>>>>> F) which makes G true in F. >>>>>>>>>>>>>>>>>>>>>>>>>>
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G exists >>>>>>>>>>>>>>>>>>>>>>>> in F.
Why do you say that?
I don't think you know what you terms mean. >>>>>>>>>>>>>>>>>>>>>>>
There exists a G in F such that G is true if and >>>>>>>>>>>>>>>>>>>>>>> only if G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true >>>>>>>>>>>>>>>>>>>>>> if and only if G is
unprovable in F making G unsatisfied thus untrue >>>>>>>>>>>>>>>>>>>>>> in F.
Row(4) There exists a G in F such that G is false >>>>>>>>>>>>>>>>>>>>>> if and only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then ↔ >>>>>>>>>>>>>>>>>>>>>> is false.
But if neither row values can ACTUALLY EXIST, then >>>>>>>>>>>>>>>>>>>>> the equality is true.
(for whatever reason) then ↔ is unsatisfied and no >>>>>>>>>>>>>>>>>>>> such G exists in F.
You don't need to have all the rows with true being >>>>>>>>>>>>>>>>>>> possible, you need all the rows that are possible to >>>>>>>>>>>>>>>>>>> be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that >>>>>>>>>>>>>>>>>> the LHS and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>>>>>>>>
Right, and for that statement, the actual G found in F, >>>>>>>>>>>>>>>>> the ONLY values that happen is G is ALWAYS true, an >>>>>>>>>>>>>>>>> Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is true and see how that >>>>>>>>>>>>>>>> effects the LHS
We must assume that the RHS is false and see how that >>>>>>>>>>>>>>>> effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡
(RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS) >>>>>>>>>>>>>>>>
Nope, that isn't how it works.
Can you show me something that says that is how it works? >>>>>>>>>>>>>>
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in >>>>>>>>>>>>> this case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules of >>>>>>>>>>>> logic the
above expression is simply false. There is no G that is >>>>>>>>>>>> logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a >>>>>>>>>>> number g such that g statisifes a particular Primative
Recursive Relationship (built in Meta-F, but using only
operations defined in F).
falsely place
the blame on F.
Yes, but can you PROVE your statement? If not, you are just
making unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of
understanding (an honest mistake not a lie) about how ↔ works. >>>>>>>>
Nope, how did you prove that no such G exists? You claims that
row 4 can't be satisfied? it doesn't need to ever be used.
Try and prove that with a source, in the mean time I will tentatively >>>>>> assume that you are wrong. I proved that I am correct with the above >>>>>> truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p)) >>>>>>
WRONG, YOU are making the claim, so YOU need to prove it.
truth table needed to be satisfied. Furthermore in retrospect this
looks
like a dumb mistake that I did not notice as a dumb mistake until I
looked at the truth table for ∧. So we are back to row one.
∃G ∈ F (G ↔ (F ⊬ G))
If the RHS is satisfied then this means that there are no inference
steps in F that derive G, thus G cannot be shown to be true in F.
Nope, there is no FINITE series of infernece steps in F that derive G.
This G cannot be shown to be true in F.
It can't be PROVEN in F, but it can be PROVEN to be true in F with a
proof in Meta-F
You just don't seem to understand how these Meta-systems work.
On 4/22/2023 4:04 PM, Richard Damon wrote:
On 4/22/23 4:47 PM, olcott wrote:
On 4/22/2023 3:43 PM, Richard Damon wrote:
On 4/22/23 4:39 PM, olcott wrote:
On 4/22/2023 3:14 PM, Richard Damon wrote:
On 4/22/23 4:10 PM, olcott wrote:I may have been mistaken when I thought that more than one row of the >>>>> truth table needed to be satisfied. Furthermore in retrospect this
On 4/22/2023 3:06 PM, Richard Damon wrote:
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not >>>>>>>>>>> falsely place
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:p ↔ q would seem to mean ((p → q) ∧ (q → p)) >>>>>>>>>>>>>>> Here is a much clearer and conventional way of showing that >>>>>>>>>>>>>>>
On 4/22/2023 11:56 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>> On 4/22/23 12:45 PM, olcott wrote:
I don't think that is the way that it works. >>>>>>>>>>>>>>>>> We must assume that the RHS is true and see how that >>>>>>>>>>>>>>>>> effects the LHSOn 4/22/2023 11:36 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 4/22/23 11:39 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>So, you don't understand how truth tables work. >>>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) cannot have the same >>>>>>>>>>>>>>>>>>>>> value for p and qBoy are you wrong.Doesn't seem so, you don't seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand the difference. You seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>There exists a G such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> F that means that* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is true means there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not exist >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F*
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that isn't what G is, you only >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> think that because you can't actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand even the outline of Godel's >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof, so you take pieces out of context. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F} >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> conventional way. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a statement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that asserts its own unprovablity, as a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> simplification describing a statment >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> DERIVED from G, and that derivation >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> happens in Meta-F, and is about what can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> be proven in F.
So?
Since Godel's G isn't of that form, but >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> only can be used to derive a statment IN >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> META-F that says that G is not provable >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F, your argument says nothing about >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Godel's G.
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to
Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Did you read that article? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Also, you don't understand what those >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> terms mean, because G being true doesn't >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mean there is no sequence of inference >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> steps that satisfies G in F, but there >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is no FINITE sequence of inference steps >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because we can see that every finite or >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> LHS a powerful F can infer that G >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequences in this more >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> powerful F.
Nope. Show the PROOF. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't know HOW to do a proof, you can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> only do arguement. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F
A proof is any sequence of steps that shows >>>>>>>>>>>>>>>>>>>>>>>>>>>>> that its conclusion is a >>>>>>>>>>>>>>>>>>>>>>>>>>>>> necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that >>>>>>>>>>>>>>>>>>>>>>>>>>>> shows that a given statement is a necessary >>>>>>>>>>>>>>>>>>>>>>>>>>>> consequence of the defined system. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
"Proof" doesn't have a "Premise", it has a >>>>>>>>>>>>>>>>>>>>>>>>>>>> system.
The statement may have conditions in it >>>>>>>>>>>>>>>>>>>>>>>>>>>> restricting when
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>> making G untrue.
no FINITE sequence, making G UNPROVABLE, and >>>>>>>>>>>>>>>>>>>>>>>>>>>> there IS an INFINITE sequence making it TRUE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>> making G true.
If G is false, then there is a finite >>>>>>>>>>>>>>>>>>>>>>>>>>>> sequence proving G, which forces G to be >>>>>>>>>>>>>>>>>>>>>>>>>>>> true, thus this is a contradiction. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS >>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no such G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>> Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite >>>>>>>>>>>>>>>>>>>>>>>>>>>> sequence that isn't finite, G can be True >>>>>>>>>>>>>>>>>>>>>>>>>>>> but not Provable.
If G is false and ↔ is true this makes the >>>>>>>>>>>>>>>>>>>>>>>>>>> RHS false which negates the RHS making it say >>>>>>>>>>>>>>>>>>>>>>>>>>> (G ⊢ F) which makes G true in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G >>>>>>>>>>>>>>>>>>>>>>>>> exists in F.
Why do you say that?
I don't think you know what you terms mean. >>>>>>>>>>>>>>>>>>>>>>>>
There exists a G in F such that G is true if and >>>>>>>>>>>>>>>>>>>>>>>> only if G is Unprovable.
Logical equality
p q p ↔ q
T T T // G is true if and only if G is Unprovable. >>>>>>>>>>>>>>>>>>>>>>> T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true >>>>>>>>>>>>>>>>>>>>>>> if and only if G is
unprovable in F making G unsatisfied thus untrue >>>>>>>>>>>>>>>>>>>>>>> in F.
Row(4) There exists a G in F such that G is false >>>>>>>>>>>>>>>>>>>>>>> if and only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then ↔ >>>>>>>>>>>>>>>>>>>>>>> is false.
But if neither row values can ACTUALLY EXIST, then >>>>>>>>>>>>>>>>>>>>>> the equality is true.
(for whatever reason) then ↔ is unsatisfied and no >>>>>>>>>>>>>>>>>>>>> such G exists in F.
You don't need to have all the rows with true being >>>>>>>>>>>>>>>>>>>> possible, you need all the rows that are possible to >>>>>>>>>>>>>>>>>>>> be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that >>>>>>>>>>>>>>>>>>> the LHS and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>>>>>>>>>
Right, and for that statement, the actual G found in >>>>>>>>>>>>>>>>>> F, the ONLY values that happen is G is ALWAYS true, an >>>>>>>>>>>>>>>>>> Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is false and see how that >>>>>>>>>>>>>>>>> effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡
(RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS) >>>>>>>>>>>>>>>>>
Nope, that isn't how it works.
Can you show me something that says that is how it works? >>>>>>>>>>>>>>>
Logical implication derives logical equivalence
p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in >>>>>>>>>>>>>> this case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules >>>>>>>>>>>>> of logic the
above expression is simply false. There is no G that is >>>>>>>>>>>>> logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a >>>>>>>>>>>> number g such that g statisifes a particular Primative >>>>>>>>>>>> Recursive Relationship (built in Meta-F, but using only >>>>>>>>>>>> operations defined in F).
the blame on F.
Yes, but can you PROVE your statement? If not, you are just >>>>>>>>>> making unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of >>>>>>>>> understanding (an honest mistake not a lie) about how ↔ works. >>>>>>>>>
Nope, how did you prove that no such G exists? You claims that >>>>>>>> row 4 can't be satisfied? it doesn't need to ever be used.
Try and prove that with a source, in the mean time I will
tentatively
assume that you are wrong. I proved that I am correct with the above >>>>>>> truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p)) >>>>>>>
WRONG, YOU are making the claim, so YOU need to prove it.
looks
like a dumb mistake that I did not notice as a dumb mistake until I
looked at the truth table for ∧. So we are back to row one.
∃G ∈ F (G ↔ (F ⊬ G))
If the RHS is satisfied then this means that there are no inference
steps in F that derive G, thus G cannot be shown to be true in F.
Nope, there is no FINITE series of infernece steps in F that derive G. >>>>
This G cannot be shown to be true in F.
It can't be PROVEN in F, but it can be PROVEN to be true in F with a
proof in Meta-F
You just don't seem to understand how these Meta-systems work.
If G cannot be satisfied in F then G is not true in F.
On 4/22/23 5:10 PM, olcott wrote:
On 4/22/2023 4:04 PM, Richard Damon wrote:
On 4/22/23 4:47 PM, olcott wrote:
On 4/22/2023 3:43 PM, Richard Damon wrote:
On 4/22/23 4:39 PM, olcott wrote:
On 4/22/2023 3:14 PM, Richard Damon wrote:
On 4/22/23 4:10 PM, olcott wrote:I may have been mistaken when I thought that more than one row of the >>>>>> truth table needed to be satisfied. Furthermore in retrospect this >>>>>> looks
On 4/22/2023 3:06 PM, Richard Damon wrote:
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not >>>>>>>>>>>> falsely place
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:p ↔ q would seem to mean ((p → q) ∧ (q → p)) >>>>>>>>>>>>>>>> Here is a much clearer and conventional way of showing that >>>>>>>>>>>>>>>>
On 4/22/2023 11:56 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 4/22/23 12:45 PM, olcott wrote:
I don't think that is the way that it works. >>>>>>>>>>>>>>>>>> We must assume that the RHS is true and see how that >>>>>>>>>>>>>>>>>> effects the LHSOn 4/22/2023 11:36 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 4/22/23 12:27 PM, olcott wrote:
On 4/22/2023 11:12 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 11:39 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>So, you don't understand how truth tables work. >>>>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) cannot have the same >>>>>>>>>>>>>>>>>>>>>> value for p and qBoy are you wrong.Doesn't seem so, you don't seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand the difference. You seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>There exists a G such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F that means that* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is true means there is no sequence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> of inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is false means there is a sequence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> of inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Thus the above G simply does not >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> exist in F* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> correctly link provability to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory way. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that isn't what G is, you only >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> think that because you can't actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand even the outline of Godel's >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof, so you take pieces out of context. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> own unprovability in F} >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> using conventional math symbols in their >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> conventional way. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> statement that asserts its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovablity, as a simplification >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> describing a statment DERIVED from G, and >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that derivation happens in Meta-F, and is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> about what can be proven in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
Since Godel's G isn't of that form, but >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> only can be used to derive a statment >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> IN META-F that says that G is not >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> provable in F, your argument says >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> nothing about Godel's G. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to
Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Did you read that article? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Also, you don't understand what those >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> terms mean, because G being true >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> doesn't mean there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> but there is no FINITE sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because we can see that every finite or >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> LHS a powerful F can infer that G >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequences in this more >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> powerful F.
Nope. Show the PROOF. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't know HOW to do a proof, you can >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> only do arguement. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> shows that its conclusion is a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that >>>>>>>>>>>>>>>>>>>>>>>>>>>>> shows that a given statement is a necessary >>>>>>>>>>>>>>>>>>>>>>>>>>>>> consequence of the defined system. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
"Proof" doesn't have a "Premise", it has a >>>>>>>>>>>>>>>>>>>>>>>>>>>>> system.
The statement may have conditions in it >>>>>>>>>>>>>>>>>>>>>>>>>>>>> restricting when
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> making G untrue.
no FINITE sequence, making G UNPROVABLE, >>>>>>>>>>>>>>>>>>>>>>>>>>>>> and there IS an INFINITE sequence making it >>>>>>>>>>>>>>>>>>>>>>>>>>>>> TRUE.
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> making G true.
If G is false, then there is a finite >>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequence proving G, which forces G to be >>>>>>>>>>>>>>>>>>>>>>>>>>>>> true, thus this is a contradiction. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no such G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite >>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequence that isn't finite, G can be True >>>>>>>>>>>>>>>>>>>>>>>>>>>>> but not Provable.
If G is false and ↔ is true this makes the >>>>>>>>>>>>>>>>>>>>>>>>>>>> RHS false which negates the RHS making it >>>>>>>>>>>>>>>>>>>>>>>>>>>> say (G ⊢ F) which makes G true in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G >>>>>>>>>>>>>>>>>>>>>>>>>> exists in F.
Why do you say that?
I don't think you know what you terms mean. >>>>>>>>>>>>>>>>>>>>>>>>>
There exists a G in F such that G is true if >>>>>>>>>>>>>>>>>>>>>>>>> and only if G is Unprovable. >>>>>>>>>>>>>>>>>>>>>>>>>
Logical equality
p q p ↔ q
T T T // G is true if and only if G is >>>>>>>>>>>>>>>>>>>>>>>> Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is Provable. >>>>>>>>>>>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is true >>>>>>>>>>>>>>>>>>>>>>>> if and only if G is
unprovable in F making G unsatisfied thus untrue >>>>>>>>>>>>>>>>>>>>>>>> in F.
Row(4) There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>> false if and only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then >>>>>>>>>>>>>>>>>>>>>>>> ↔ is false.
But if neither row values can ACTUALLY EXIST, >>>>>>>>>>>>>>>>>>>>>>> then the equality is true.
(for whatever reason) then ↔ is unsatisfied and no >>>>>>>>>>>>>>>>>>>>>> such G exists in F.
You don't need to have all the rows with true being >>>>>>>>>>>>>>>>>>>>> possible, you need all the rows that are possible >>>>>>>>>>>>>>>>>>>>> to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that >>>>>>>>>>>>>>>>>>>> the LHS and the RHS
must always have the same truth value or ↔ is not true. >>>>>>>>>>>>>>>>>>>>
Right, and for that statement, the actual G found in >>>>>>>>>>>>>>>>>>> F, the ONLY values that happen is G is ALWAYS true, >>>>>>>>>>>>>>>>>>> an Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is false and see how that >>>>>>>>>>>>>>>>>> effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) >>>>>>>>>>>>>>>>>> ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS) >>>>>>>>>>>>>>>>>>
Nope, that isn't how it works.
Can you show me something that says that is how it works? >>>>>>>>>>>>>>>>
Logical implication derives logical equivalence >>>>>>>>>>>>>>>> p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used in >>>>>>>>>>>>>>> this case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules >>>>>>>>>>>>>> of logic the
above expression is simply false. There is no G that is >>>>>>>>>>>>>> logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a >>>>>>>>>>>>> number g such that g statisifes a particular Primative >>>>>>>>>>>>> Recursive Relationship (built in Meta-F, but using only >>>>>>>>>>>>> operations defined in F).
the blame on F.
Yes, but can you PROVE your statement? If not, you are just >>>>>>>>>>> making unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of >>>>>>>>>> understanding (an honest mistake not a lie) about how ↔ works. >>>>>>>>>>
Nope, how did you prove that no such G exists? You claims that >>>>>>>>> row 4 can't be satisfied? it doesn't need to ever be used.
Try and prove that with a source, in the mean time I will
tentatively
assume that you are wrong. I proved that I am correct with the >>>>>>>> above
truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p)) >>>>>>>>
WRONG, YOU are making the claim, so YOU need to prove it.
like a dumb mistake that I did not notice as a dumb mistake until I >>>>>> looked at the truth table for ∧. So we are back to row one.
∃G ∈ F (G ↔ (F ⊬ G))
If the RHS is satisfied then this means that there are no
inference steps in F that derive G, thus G cannot be shown to be
true in F.
Nope, there is no FINITE series of infernece steps in F that derive G. >>>>>
This G cannot be shown to be true in F.
It can't be PROVEN in F, but it can be PROVEN to be true in F with a
proof in Meta-F
You just don't seem to understand how these Meta-systems work.
If G cannot be satisfied in F then G is not true in F.
Who says G can not be satisified in F?
In fact, Godel's G has no model variables in it that need to be
satisfied. G is just UNCONDITIONALLY TRUE in all the models of F.
I don't think you actually understand what you are talking about.
That just shows how ignorant you are.
On 4/22/2023 4:26 PM, Richard Damon wrote:
On 4/22/23 5:10 PM, olcott wrote:
On 4/22/2023 4:04 PM, Richard Damon wrote:
On 4/22/23 4:47 PM, olcott wrote:
On 4/22/2023 3:43 PM, Richard Damon wrote:
On 4/22/23 4:39 PM, olcott wrote:
On 4/22/2023 3:14 PM, Richard Damon wrote:
On 4/22/23 4:10 PM, olcott wrote:I may have been mistaken when I thought that more than one row of >>>>>>> the
On 4/22/2023 3:06 PM, Richard Damon wrote:
On 4/22/23 4:02 PM, olcott wrote:
On 4/22/2023 3:00 PM, Richard Damon wrote:
On 4/22/23 3:54 PM, olcott wrote:
On 4/22/2023 2:44 PM, Richard Damon wrote:
On 4/22/23 3:34 PM, olcott wrote:There is no such G in F says the same thing, yet does not >>>>>>>>>>>>> falsely place
On 4/22/2023 2:15 PM, Richard Damon wrote:
On 4/22/23 3:11 PM, olcott wrote:
On 4/22/2023 1:01 PM, Richard Damon wrote:
On 4/22/23 1:13 PM, olcott wrote:p ↔ q would seem to mean ((p → q) ∧ (q → p)) >>>>>>>>>>>>>>>>> Here is a much clearer and conventional way of showing >>>>>>>>>>>>>>>>> that
On 4/22/2023 11:56 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 4/22/23 12:45 PM, olcott wrote:
I don't think that is the way that it works. >>>>>>>>>>>>>>>>>>> We must assume that the RHS is true and see how that >>>>>>>>>>>>>>>>>>> effects the LHSOn 4/22/2023 11:36 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 4/22/23 12:27 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 11:12 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 11:39 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
So, you don't understand how truth tables work. >>>>>>>>>>>>>>>>>>>>>>If either Row(1) or Row(4) cannot have the same >>>>>>>>>>>>>>>>>>>>>>> value for p and qBoy are you wrong. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>There exists a G such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*If we assume that there is such a G >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F that means that* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> G is true means there is no sequence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> of inference steps that satisfies G >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F.
G is false means there is a sequence >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> of inference steps that satisfies G >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> in F.
*Thus the above G simply does not >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> exist in F* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
I finally learned enough model theory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to correctly link provability to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> truth in the conventional model theory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> way.
Doesn't seem so, you don't seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand the difference. You seem to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> confuse Truth with Knowledge. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that isn't what G is, you only >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> think that because you can't actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> understand even the outline of Godel's >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> proof, so you take pieces out of context. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I finally approximated {G asserts its >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> own unprovability in F} >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> using conventional math symbols in >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> their conventional way. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
G never asserts its own unprovability. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The statement that we now have a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> statement that asserts its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovablity, as a simplification >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> describing a statment DERIVED from G, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and that derivation happens in Meta-F, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and is about what can be proven in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
So?
Since Godel's G isn't of that form, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> but only can be used to derive a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> statment IN META-F that says that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> not provable in F, your argument says >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> nothing about Godel's G. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
F ⊢ GF ↔ ¬ProvF (┌GF┐). >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
I have finally created a G that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> equivalent to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Did you read that article? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Also, you don't understand what those >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> terms mean, because G being true >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> doesn't mean there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> but there is no FINITE sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because we can see that every finite or >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequence in F that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> LHS a powerful F can infer that G >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is utterly unsatisfiable even for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> infinite sequences in this more >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> powerful F.
Nope. Show the PROOF. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't know HOW to do a proof, you >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> can only do arguement. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A proof is any sequence of steps that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> shows that its conclusion is a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> necessary consequence of its premises.\ >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
A proof is a FINITE sequence of steps that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> shows that a given statement is a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> necessary consequence of the defined system. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
"Proof" doesn't have a "Premise", it has >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a system.
The statement may have conditions in it >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> restricting when
∃G ∈ F (G ↔ (G ⊬ F)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> logically equivalent to its own >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unprovability in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
If G is true then there is no sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> making G untrue.
no FINITE sequence, making G UNPROVABLE, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and there IS an INFINITE sequence making >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> it TRUE.
This is possible.
If G is false then there is a sequence of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> inference steps that satisfies G in F >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> making G true.
If G is false, then there is a finite >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequence proving G, which forces G to be >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> true, thus this is a contradiction. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Because the RHS of ↔ contradicts the LHS >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no such G in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Thus the above G simply does not exist in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, because we can have an infinite >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> sequence that isn't finite, G can be True >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> but not Provable.
If G is false and ↔ is true this makes the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> RHS false which negates the RHS making it >>>>>>>>>>>>>>>>>>>>>>>>>>>>> say (G ⊢ F) which makes G true in F. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Right, G can't be false, but it can be True. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Thus ↔ cannot be satisfied thus no such G >>>>>>>>>>>>>>>>>>>>>>>>>>> exists in F.
Why do you say that?
I don't think you know what you terms mean. >>>>>>>>>>>>>>>>>>>>>>>>>>
There exists a G in F such that G is true if >>>>>>>>>>>>>>>>>>>>>>>>>> and only if G is Unprovable. >>>>>>>>>>>>>>>>>>>>>>>>>>
Logical equality
p q p ↔ q
T T T // G is true if and only if G is >>>>>>>>>>>>>>>>>>>>>>>>> Unprovable.
T F F //
F T F //
F F T // G is false if and only if G is >>>>>>>>>>>>>>>>>>>>>>>>> Provable.
https://en.wikipedia.org/wiki/Truth_table#Logical_equality
Row(1) There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>> true if and only if G is
unprovable in F making G unsatisfied thus >>>>>>>>>>>>>>>>>>>>>>>>> untrue in F.
Row(4) There exists a G in F such that G is >>>>>>>>>>>>>>>>>>>>>>>>> false if and only if G is
provable in F making G satisfied thus true in F. >>>>>>>>>>>>>>>>>>>>>>>>>
If either Row(1) or Row(4) are unsatisfied then >>>>>>>>>>>>>>>>>>>>>>>>> ↔ is false.
But if neither row values can ACTUALLY EXIST, >>>>>>>>>>>>>>>>>>>>>>>> then the equality is true.
(for whatever reason) then ↔ is unsatisfied and >>>>>>>>>>>>>>>>>>>>>>> no such G exists in F.
You don't need to have all the rows with true >>>>>>>>>>>>>>>>>>>>>> being possible, you need all the rows that are >>>>>>>>>>>>>>>>>>>>>> possible to be True.
To the best of my knowledge
↔ is also known as logical equivalence meaning that >>>>>>>>>>>>>>>>>>>>> the LHS and the RHS
must always have the same truth value or ↔ is not >>>>>>>>>>>>>>>>>>>>> true.
Right, and for that statement, the actual G found in >>>>>>>>>>>>>>>>>>>> F, the ONLY values that happen is G is ALWAYS true, >>>>>>>>>>>>>>>>>>>> an Unprovable is always true.
Thus the equivalence is always true.
We must assume that the RHS is false and see how that >>>>>>>>>>>>>>>>>>> effects the LHS
((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) >>>>>>>>>>>>>>>>>>> ≡ (RHS ↔ LHS)
False(RHS) → True(LHS) refutes (RHS ↔ LHS) >>>>>>>>>>>>>>>>>>>
Nope, that isn't how it works.
Can you show me something that says that is how it works? >>>>>>>>>>>>>>>>>
Logical implication derives logical equivalence >>>>>>>>>>>>>>>>> p---q---(p ⇒ q)---(q ⇒ p)---(q ↔ p)
T---T------T----------T---------T
T---F------F----------T---------F
F---T------T----------F---------F
F---F------T----------T---------T
So, why does the fact that the last line is never used >>>>>>>>>>>>>>>> in this case cause a problem.
∃G ∈ F (G ↔ (G ⊬ F))
I am just saying that according to the conventional rules >>>>>>>>>>>>>>> of logic the
above expression is simply false. There is no G that is >>>>>>>>>>>>>>> logically
equivalent to its own unprovability in F.
But Godel's G satisfies that.
Remember, G is the statement that there does not exist a >>>>>>>>>>>>>> number g such that g statisifes a particular Primative >>>>>>>>>>>>>> Recursive Relationship (built in Meta-F, but using only >>>>>>>>>>>>>> operations defined in F).
the blame on F.
Yes, but can you PROVE your statement? If not, you are just >>>>>>>>>>>> making unsubstantiated false claims, just like DT.
I just proved it. The only gap in the proof was your lack of >>>>>>>>>>> understanding (an honest mistake not a lie) about how ↔ works. >>>>>>>>>>>
Nope, how did you prove that no such G exists? You claims that >>>>>>>>>> row 4 can't be satisfied? it doesn't need to ever be used.
Try and prove that with a source, in the mean time I will
tentatively
assume that you are wrong. I proved that I am correct with the >>>>>>>>> above
truth table yet this assumes: p ↔ q means ((p → q) ∧ (q → p)) >>>>>>>>>
WRONG, YOU are making the claim, so YOU need to prove it.
truth table needed to be satisfied. Furthermore in retrospect
this looks
like a dumb mistake that I did not notice as a dumb mistake until I >>>>>>> looked at the truth table for ∧. So we are back to row one.
∃G ∈ F (G ↔ (F ⊬ G))
If the RHS is satisfied then this means that there are no
inference steps in F that derive G, thus G cannot be shown to be >>>>>>> true in F.
Nope, there is no FINITE series of infernece steps in F that
derive G.
This G cannot be shown to be true in F.
It can't be PROVEN in F, but it can be PROVEN to be true in F with a
proof in Meta-F
You just don't seem to understand how these Meta-systems work.
If G cannot be satisfied in F then G is not true in F.
Who says G can not be satisified in F?
To derive G in F requires a set of inference steps in F that proves that these same inference steps do not exist in F.
In fact, Godel's G has no model variables in it that need to be
satisfied. G is just UNCONDITIONALLY TRUE in all the models of F.
I don't think you actually understand what you are talking about.
That just shows how ignorant you are.
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