When PA is the n level of logic and metamathematics is the n+1 level of
logic in the same formal system then G <is> provable in this formal
system and incompleteness cannot exist. HOL can do this.
This sentence is not true: "This sentence is not true" is true.
The above is all there is to the Tarski Undefinability Theorem. https://liarparadox.org/Tarski_275_276.pdf
On 12/2/2023 6:22 PM, olcott wrote:
When PA is the n level of logic and metamathematics is the n+1 level of
logic in the same formal system then G <is> provable in this formal
system and incompleteness cannot exist. HOL can do this.
This sentence is not true: "This sentence is not true" is true.
The above is all there is to the Tarski Undefinability Theorem.
https://liarparadox.org/Tarski_275_276.pdf
When two formal systems are included in the same formal system
as logic levels of n and n+1 then anything expressed in the
n level of PA is provable in the n+1 level of metamathematics
and incompleteness cannot exist.
If G is expressed at the n+1 level then the n+2 level proves G.
On 12/2/2023 9:34 PM, olcott wrote:
On 12/2/2023 6:22 PM, olcott wrote:
When PA is the n level of logic and metamathematics is the n+1 level of
logic in the same formal system then G <is> provable in this formal
system and incompleteness cannot exist. HOL can do this.
This sentence is not true: "This sentence is not true" is true.
The above is all there is to the Tarski Undefinability Theorem.
https://liarparadox.org/Tarski_275_276.pdf
When two formal systems are included in the same formal system
as logic levels of n and n+1 then anything expressed in the
n level of PA is provable in the n+1 level of metamathematics
and incompleteness cannot exist.
If G is expressed at the n+1 level then the n+2 level proves G.
A formal system having an unlimited number of logic levels cannot
possibly be incomplete in the Gödel sense. Whatsoever can be
expressed at the n level of logic can be proven at the n+1 level
of logic of this same formal system.
When PA is the n level of logic and
metamathematics is the n+1 level of logic
in the same formal system
then
G <is> provable in this formal system
and incompleteness cannot exist.
HOL can do this.
This sentence is not true:
"This sentence is not true"
is true.
The above is all there is to
the Tarski Undefinability Theorem.
https://liarparadox.org/Tarski_275_276.pdf
On 12/2/2023 10:23 PM, olcott wrote:
On 12/2/2023 9:34 PM, olcott wrote:
On 12/2/2023 6:22 PM, olcott wrote:
When PA is the n level of logic and metamathematics is the n+1 level of >>>> logic in the same formal system then G <is> provable in this formal
system and incompleteness cannot exist. HOL can do this.
This sentence is not true: "This sentence is not true" is true.
The above is all there is to the Tarski Undefinability Theorem.
https://liarparadox.org/Tarski_275_276.pdf
When two formal systems are included in the same formal system
as logic levels of n and n+1 then anything expressed in the
n level of PA is provable in the n+1 level of metamathematics
and incompleteness cannot exist.
If G is expressed at the n+1 level then the n+2 level proves G.
A formal system having an unlimited number of logic levels cannot
possibly be incomplete in the Gödel sense. Whatsoever can be
expressed at the n level of logic can be proven at the n+1 level
of logic of this same formal system.
HOL can have from 0 to N contiguous logic levels where N is a natural
number, yet not a fixed constant.
On 12/3/2023 6:46 AM, Jim Burns wrote:
On 12/2/2023 7:22 PM, olcott wrote:
When PA is the n level of logic and
metamathematics is the n+1 level of logic
in the same formal system
then
G <is> provable in this formal system
and incompleteness cannot exist.
HOL can do this.
One PA.sentence not PA.provable
makes PA incomplete.
One PA.sentence PA.provable
does not make PA complete.
One PA.sentence HOL.provable
does not make PA complete.
One PA.sentence HOL.provable
does not make HOL complete.
One HOL.sentence HOL.provable
does not make HOL complete.
One HOL.sentence not HOL.provable
makes HOL incomplete.
HOL can have
from 0 to N contiguous logic levels
where N is a natural number,
yet not a fixed constant.
This sentence cannot be proved:
"This sentence cannot be proven" can be proved.
HOL always has
as many orders of logic needed so that
anything that can be specified
can be proved.
On 12/3/2023 11:10 AM, Jim Burns wrote:
[...]
The sentence that you provided
is simply rejected as incoherent.
HOL.{This sentence cannot be proven}.
on the basis that it has
a cycle in its evaluation directed graph.
The version that you provided
is simply
a screwy way to indirectly refer to
a sentence that is isomorphic to
referring to its by its variable name.
On 12/3/2023 6:46 AM, Jim Burns wrote:
On 12/2/2023 7:22 PM, olcott wrote:
When PA is the n level of logic and
metamathematics is the n+1 level of logic
in the same formal system
then
G <is> provable in this formal system
and incompleteness cannot exist.
HOL can do this.
One PA.sentence not PA.provable
makes PA incomplete.
One PA.sentence PA.provable
does not make PA complete.
One PA.sentence HOL.provable
does not make PA complete.
One PA.sentence HOL.provable
does not make HOL complete.
One HOL.sentence HOL.provable
does not make HOL complete.
One HOL.sentence not HOL.provable
makes HOL incomplete.
HOL can have from 0 to N contiguous logic levels where N
is a natural number, yet not a fixed constant.
This sentence cannot be proved: "This sentence cannot be proven"
can be proved.
HOL always has as many orders of logic needed so that anything
that can be specified can be proved.
On 12/3/2023 1:03 PM, Jim Burns wrote:
On 12/3/2023 12:32 PM, olcott wrote:
The version that you provided
This one HOL.sentence

 HOL" preceded by its HOL.quotation
 is not HOL.provable "HOL
 preceded by its HOL.quotation
 is not HOL.provable.
Is rejected as semantically unsound
on the basis that
it forms a cycle in the directed graph of
its evaluation sequence.
"preceded by its HOL.quotation" is simply
a way to indirectly refer to a sentence
by its name.
On 12/3/2023 4:14 PM, Jim Burns wrote:
We need to be able to express two properties
in our languageoftheday for HOL
Arithmetic is sufficient for this, but
we can go further, as long as
we can still express it.
Proves(y,x)
if
y represents a proof in HOL of
a formula in HOL which x represents
then
Proves(y,x) is true. Otherwise, false.
Subst(x,y,z)
if
x represents a formula A(u) with
one variable not bound by a quantifier (free)
and z represents A(y)
AKA A(u) with y substituted for
free occurrences of u in A(u)
then
Subst(x,y,z) is true. Otherwise, false.
Whenever any proof contains cycles in
its directed graph
it is semantically unsound and must be rejected.
On 12/3/2023 9:57 PM, Jim Burns wrote:
[...]
https://en.wikipedia.org/wiki/Quine%27s_paradox
A convoluted mess that
is indirectly rather than directly
selfreferential.
 The statement
 "'yields falsehood when preceded by its quotation'
 yields falsehood when preceded by its quotation"
 is false.
 In other words,
 the sentence implies that it is false,
 which is paradoxical
 —for if it is false,
 what it states is in fact true.
*Thus proving a cycle in its evaluation graph*
All proves are directed paths
from leaves to a root.
On 12/4/2023 8:15 AM, Jim Burns wrote:
On 12/4/2023 12:01 AM, olcott wrote:
On 12/3/2023 9:57 PM, Jim Burns wrote:
[...]
https://en.wikipedia.org/wiki/Quine%27s_paradox
A convoluted mess that
is indirectly rather than directly
selfreferential.
 The statement
 "'yields falsehood when preceded by its quotation'
 yields falsehood when preceded by its quotation"
 is false.
G(x) :⇔ ∃y(Subst(x,x,y) ∧ ¬∃z:STProves(z,y)))
ST is: Empty.Set, Adjunct, Extensionality.
h := "H(x)"
G(h) means
 H(x) preceded by its quotation
 is not ST.provable.
g := "G(x)" ≠ "G(g)"
G(g) means
 G(x) preceded by its quotation
 is not ST.provable.
!
G(g) is G(x) preceded by its quotation.
or,
being cute,
 "preceded by its quotation
 is not ST.provable"
 preceded by its quotation
 is not ST.provable.
If G(g) is true, G(g) is not ST.provable.
g refers to G(x) not G(g)
 In other words,
 the sentence implies that it is false,
G(g) implies that G(g) is not ST.provable.
false ≠ not ST.provable
 which is paradoxical
 —for if it is false,
 what it states is in fact true.
Incomplete ST is not paradoxical,
not even if you say it is
a hundred times.
*Thus proving a cycle in its evaluation graph*
G(g)
Show me a cycle.
All proves are directed paths
from leaves to a root.
...which is what I've been saying,
calling the root
description of the topicoftheday and
calling the leaves
notfirstfalse claims augmenting
the description.
The finiteness of paths from leaves
back to root is implicit in
their being attached.
One can make any idea sufficiently convoluted
to make it incomprehensible.
When one knows the foundation of formal proofs always
involves a directed path from the leaves of a tree
to its root then one has the basis to untangle the
fundamental nature of any formal proof once it has
been translated into its directed graph form.
...14 Every epistemological antinomy can likewise be used
for a similar undecidability proof...(Gödel 1931:4344)
When it is understood that an epistemological antinomy
is merely a selfcontradictory expression of language
then the above quote conclusively proves that Gödel had a
fundamental misconception about the way the formal proofs
actually work.
It is far too ridiculously stupid to believe that anyone
ever truly believed that a formal system must be able
to prove any selfcontradictory expression of language.
Thus the Gödel proof becomes the Gödel ruse.
On 12/4/2023 8:15 AM, Jim Burns wrote:
On 12/4/2023 12:01 AM, olcott wrote:
*Thus proving a cycle in its evaluation graph*
G(g)
Show me a cycle.
One can make any idea sufficiently convoluted
to make it incomprehensible.
On 12/2/2023 6:22 PM, olcott wrote:
When PA is the n level of logic and metamathematics is the n+1 level of
logic in the same formal system then G <is> provable in this formal
system and incompleteness cannot exist. HOL can do this.
This sentence is not true: "This sentence is not true" is true.
The above is all there is to the Tarski Undefinability Theorem.
https://liarparadox.org/Tarski_275_276.pdf
...14 Every epistemological antinomy can likewise be used
for a similar undecidability proof...(Gödel 1931:4344)
Dishonest people can "interpret" this as saying everyone
knows that no one can ever base any formal proof on any
epistemological antinomy.
*the direct opposite of what he actually says*
On 12/4/23 8:00 PM, olcott wrote:
On 12/4/2023 6:49 PM, Jim Burns wrote:
tl;dr
if G("G(u)") is true
then G("G(u)") is not ST.provable
Until you get down to "G is unprovable in F"
we cannot build its proof tree.
Except that since that ISN'T the statment of G,
we don't need that statement.
The statement of G is:
"There does not exist
a Natural Number g that statisfies
<a particular Primative Recursive Relationship>"
There exist a proof tree for that
(which is infinite in length) in F,
since that relationship can be computed
for every Natural Number,
and it turns out that
all the answers are that
none of the numbers satisfies the relationship.
Thus, the statement IS true.
The fact that
none of the numbers satisfies the relationship
can't be determined in F, except by
testing every number individually, which,
because there is an infinite number of numbers,
says that can't be the needed finite proof.
On 12/5/2023 11:52 AM, Jim Burns wrote:
On 12/4/2023 8:44 PM, Richard Damon wrote:
On 12/4/23 8:00 PM, olcott wrote:
On 12/4/2023 6:49 PM, Jim Burns wrote:
tl;dr
if G("G(u)") is true
then G("G(u)") is not ST.provable
ST has {}, x∪{y}, and set equality.
Until you get down to "G is unprovable in F"
we cannot build its proof tree.
he sees that complete massivedatabase
as essential to True AI and
True AI as essential to humanity.
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
On 12/5/2023 1:58 PM, Jim Burns wrote:
On 12/5/2023 2:30 PM, olcott wrote:
On 12/5/2023 11:52 AM, Jim Burns wrote:
On 12/4/2023 8:44 PM, Richard Damon wrote:
On 12/4/23 8:00 PM, olcott wrote:
On 12/4/2023 6:49 PM, Jim Burns wrote:
tl;dr
if G("G(u)") is true
then G("G(u)") is not ST.provable
ST has {}, x∪{y}, and set equality.
Until you get down to
"G is unprovable in F"
we cannot build its proof tree.
What is F ?
F is the formal system that encodes Gödel’s G https://plato.stanford.edu/entries/goedelincompleteness/#FirIncTheCom
he sees that complete massivedatabase
as essential to True AI and
True AI as essential to humanity.
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
Do I have your motivation correct?
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
Eliminates Tarski undefinability and Gödel incompleteness and forces the concept of truth in math and logic to conform to the way that it works everywhere else. True(x) ≡ (⊢ x)
The only way that we know that "cats are animals" is true is the
connection from the above expression to the definition of {cats} and {animals}.
The lack of inference steps from axioms to an expression simply means
untrue. Incomplete(L) is merely a terribly misleading euphemism for ~True(L,x).
On 12/4/2023 8:44 PM, Richard Damon wrote:
On 12/4/23 8:00 PM, olcott wrote:
On 12/4/2023 6:49 PM, Jim Burns wrote:
tl;dr
if G("G(u)") is true
then G("G(u)") is not ST.provable
ST has {}, x∪{y}, and set equality.
Until you get down to "G is unprovable in F"
we cannot build its proof tree.
Except that since that ISN'T the statment of G,
we don't need that statement.
The statement of G is:
"There does not exist
a Natural Number g that statisfies
<a particular Primative Recursive Relationship>"
There exist a proof tree for that
(which is infinite in length) in F, since that relationship can be
computed
for every Natural Number,
and it turns out that
all the answers are that
none of the numbers satisfies the relationship.
Thus, the statement IS true.
The fact that
none of the numbers satisfies the relationship
can't be determined in F, except by
testing every number individually, which, because there is an infinite
number of numbers,
says that can't be the needed finite proof.
What is F ?
Is F True Arithmetic?
https://en.wikipedia.org/wiki/True_arithmetic
True Arithmetic takes infinitelymany sentences
as axioms. True Arithmetic is not a counter
example to Gödel's reasoning.
What Olcott wants still seems to be a
massivedatabase of all human knowledge,
complete, of course, and
PO sees Gödel's reasoning as what prevents that,
somehow, and
he sees that complete massivedatabase
as essential to True AI and
True AI as essential to humanity.
I do not guarantee that my recollection is
100% correct or thoroughly uptodate.
However, assuming IRC,
True Arithmetic does not serve Olcott's purpose
because
the most massively massive of massivedatabases
is still finite, unlike True Arithmetic.
However,
I'm not sure what F means here.
On 12/5/23 12:52 PM, Jim Burns wrote:
On 12/4/2023 8:44 PM, Richard Damon wrote:
On 12/4/23 8:00 PM, olcott wrote:
On 12/4/2023 6:49 PM, Jim Burns wrote:
tl;dr
if G("G(u)") is true
then G("G(u)") is not ST.provable
Until you get down to
"G is unprovable in F"
we cannot build its proof tree.
There exist a proof tree for that
(which is infinite in length) in F,
However,
I'm not sure what F means here.
F is the name of the Logic System used in
the version of Godel's paper that
Olcott is using to (mis)understand
Godel's proof.
In F,
we get a statement of the form that
G is the assertion that
there does not exist
a Natuaral Number g that satisfies
a particular Primative Recursive Relationship.
This relationship being derived in
a metasystem of F that enumerates
the fundamental properties in F,
and in which
an encoding of statements to numbers
is defined,
and the relationship is
a Proof Tester that accepts
a number (and only those numbers) that
encode
a valid proof of the statement G.
He seems to have a problem with the fact that
SOME truths can't be proven, and
somehow thinks that means that
this allows untruths to be claimed as truth.
He seems to want there to be
a massive database of all knowledge,
and then that will produce a list of
everything that is actually True and False,
and thus somehow
make lies go away.
The irony is
he atually uses the methodology of
the "fake news" crowd to try to push
his idea to be used to eliminate "fake news",
because he just doesn't understand
On 12/6/2023 11:21 AM, Jim Burns wrote:
[...]
True(x) ≡ (⊢ x)
It includes everything known to be true on
the basis of its meaning and excludes
unknown truths
As soon as humans accept
the correct measure of True(x)
then we can manually create
formal proofs of English statements.
On 12/6/2023 2:33 PM, Jim Burns wrote:is
On 12/6/2023 2:08 PM, olcott wrote:
On 12/6/2023 11:21 AM, Jim Burns wrote:
[...]
True(x) ≡ (⊢ x)
It includes everything known to be true on
the basis of its meaning and excludes
unknown truths
Unknown truths are at least as capable of
killing you as known truths.
That is something our tigerdodging ancestors
would have been able to explain well to you,
if they didn't give up and kick you out
into the dark, to let you find out for yourself,
for a short, adrenalized period of time.
As soon as humans accept
the correct measure of True(x)
then we can manually create
formal proofs of English statements.
Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
It seems that you ignored all of
my important points.
excludes unknown truths
Can you try again and ask to have
anything that you don't understand
explained?
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
That all seems to be gibberish to me.
On 12/6/2023 4:35 PM, Jim Burns wrote:
On 12/6/2023 4:55 PM, olcott wrote:
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Fascinating.
Would you like me to tell you about
global warming?
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Consider reading the post to which you respond.
qnff =
 Q is notfirst false
finseq =
 If this finite sequence of claims
 holds a false claim,
 then it holds a first false claim.
Abbreviate
a definition of "n is a natural number"
as "n is a natrual number.
natnum =
 n is a natural number
 if and only if
 n satisfies the definition of natural number

Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
Here's the problem:
You:
That all seems to be gibberish to me.
You are merely changing the subject away from the point.
My whole point pertains to True(L,x) and everything else
is off topic.
Everything about this must be able to be translated
into a clear, correct English sentence.
When G only has a meaning based on 85 instances of
indirect reference specified by 85 different math
formulas this *is not* specified clearly enough.
It must be as clear as this:
∃G ∈ WFF(F) (G ↔ (F ⊬ G))
To say that a Natural number is true or false is
nonsense and you know it.
On 12/6/2023 4:35 PM, Jim Burns wrote:
On 12/6/2023 4:55 PM, olcott wrote:
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Fascinating.
Would you like me to tell you about
global warming?
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Consider reading the post to which you respond.
qnff =
 Q is notfirst false
finseq =
 If this finite sequence of claims
 holds a false claim,
 then it holds a first false claim.
Abbreviate
a definition of "n is a natural number"
as "n is a natrual number.
natnum =
 n is a natural number
 if and only if
 n satisfies the definition of natural number

Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
Here's the problem:
You:
That all seems to be gibberish to me.
For you to understand what I am saying you must learn a little
philosophy.
The Sapir–Whorf hypothesis shows that there may be some
concepts that cannot be expressed within the scope of the
terms of logic.
https://en.wikipedia.org/wiki/Linguistic_relativity
Everything that is true on the basis of its meaning: https://en.wikipedia.org/wiki/Analytic%E2%80%93synthetic_distinction
AKA the analytic side of the analytic / synthetic distinction
necessarily must have a connection from an expression
to this meaning as its truthmaker or it cannot possibly be true.
Although within the conventional terms of logic there
may be some truths that cannot be proven there cannot
be analytic expressions of language that are true without
something making them true.
On 12/6/2023 9:56 PM, olcott wrote:
On 12/6/2023 4:35 PM, Jim Burns wrote:
On 12/6/2023 4:55 PM, olcott wrote:
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Fascinating.
Would you like me to tell you about
global warming?
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Consider reading the post to which you respond.
qnff =
 Q is notfirst false
finseq =
 If this finite sequence of claims
 holds a false claim,
 then it holds a first false claim.
Abbreviate
a definition of "n is a natural number"
as "n is a natrual number.
natnum =
 n is a natural number
 if and only if
 n satisfies the definition of natural number

Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
Here's the problem:
You:
That all seems to be gibberish to me.
For you to understand what I am saying you must learn a little
philosophy.
The Sapir–Whorf hypothesis shows that there may be some
concepts that cannot be expressed within the scope of the
terms of logic.
https://en.wikipedia.org/wiki/Linguistic_relativity
Everything that is true on the basis of its meaning:
https://en.wikipedia.org/wiki/Analytic%E2%80%93synthetic_distinction
AKA the analytic side of the analytic / synthetic distinction
necessarily must have a connection from an expression
to this meaning as its truthmaker or it cannot possibly be true.
Although within the conventional terms of logic there
may be some truths that cannot be proven there cannot
be analytic expressions of language that are true without
something making them true.
I have diligently accounted for the difference between analytical truth
and analytical knowledge the former may require an infinite sequence of
steps as its truthmaker.
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
Eliminates Tarski undefinability and Gödel incompleteness and forces the concept of truth in math and logic to conform to the way that it works everywhere else in the body of human knowledge: True(x) ≡ (⊢ x)
If the Goldbach conjecture only has an infinite sequence of steps as
its truthmaker and formal proofs do not allow an infinite sequence of
steps then we have an analytical truth with no proof yet it still has a truthmaker.
On 12/7/2023 10:20 AM, olcott wrote:
On 12/6/2023 9:56 PM, olcott wrote:
On 12/6/2023 4:35 PM, Jim Burns wrote:
On 12/6/2023 4:55 PM, olcott wrote:
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Fascinating.
Would you like me to tell you about
global warming?
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Consider reading the post to which you respond.
qnff =
 Q is notfirst false
finseq =
 If this finite sequence of claims
 holds a false claim,
 then it holds a first false claim.
Abbreviate
a definition of "n is a natural number"
as "n is a natrual number.
natnum =
 n is a natural number
 if and only if
 n satisfies the definition of natural number

Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
Here's the problem:
You:
That all seems to be gibberish to me.
For you to understand what I am saying you must learn a little
philosophy.
The Sapir–Whorf hypothesis shows that there may be some
concepts that cannot be expressed within the scope of the
terms of logic.
https://en.wikipedia.org/wiki/Linguistic_relativity
Everything that is true on the basis of its meaning:
https://en.wikipedia.org/wiki/Analytic%E2%80%93synthetic_distinction
AKA the analytic side of the analytic / synthetic distinction
necessarily must have a connection from an expression
to this meaning as its truthmaker or it cannot possibly be true.
Although within the conventional terms of logic there
may be some truths that cannot be proven there cannot
be analytic expressions of language that are true without
something making them true.
I have diligently accounted for the difference between analytical truth
and analytical knowledge the former may require an infinite sequence of
steps as its truthmaker.
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
Eliminates Tarski undefinability and Gödel incompleteness and forces
the concept of truth in math and logic to conform to the way that it
works everywhere else in the body of human knowledge: True(x) ≡ (⊢ x)
If the Goldbach conjecture only has an infinite sequence of steps as
its truthmaker and formal proofs do not allow an infinite sequence of
steps then we have an analytical truth with no proof yet it still has a
truthmaker.
That there cannot be any analytic truth without a truthmaker
refutes the Tarski Undecidability theorem.
...14 Every epistemological antinomy can likewise be used for a similar undecidability proof...(Gödel 1931:4344)
Tarski anchored his proof in an epistemological antinomy just like the
above quote: (3) x ∉ Provable if and only if x ∈ True. https://liarparadox.org/Tarski_275_276.pdf
Epistemological antinomies cannot possibly have a truthmaker (not even
with an infinite number of steps) thus are simply untrue.
On 12/7/2023 6:11 PM, olcott wrote:
On 12/7/2023 10:20 AM, olcott wrote:
On 12/6/2023 9:56 PM, olcott wrote:
On 12/6/2023 4:35 PM, Jim Burns wrote:
On 12/6/2023 4:55 PM, olcott wrote:
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Fascinating.
Would you like me to tell you about
global warming?
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Consider reading the post to which you respond.
qnff =
 Q is notfirst false
finseq =
 If this finite sequence of claims
 holds a false claim,
 then it holds a first false claim.
Abbreviate
a definition of "n is a natural number"
as "n is a natrual number.
natnum =
 n is a natural number
 if and only if
 n satisfies the definition of natural number

Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
Here's the problem:
You:
That all seems to be gibberish to me.
For you to understand what I am saying you must learn a little
philosophy.
The Sapir–Whorf hypothesis shows that there may be some
concepts that cannot be expressed within the scope of the
terms of logic.
https://en.wikipedia.org/wiki/Linguistic_relativity
Everything that is true on the basis of its meaning:
https://en.wikipedia.org/wiki/Analytic%E2%80%93synthetic_distinction
AKA the analytic side of the analytic / synthetic distinction
necessarily must have a connection from an expression
to this meaning as its truthmaker or it cannot possibly be true.
Although within the conventional terms of logic there
may be some truths that cannot be proven there cannot
be analytic expressions of language that are true without
something making them true.
I have diligently accounted for the difference between analytical truth
and analytical knowledge the former may require an infinite sequence of
steps as its truthmaker.
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
Eliminates Tarski undefinability and Gödel incompleteness and forces
the concept of truth in math and logic to conform to the way that it
works everywhere else in the body of human knowledge: True(x) ≡ (⊢ x) >>>
If the Goldbach conjecture only has an infinite sequence of steps as
its truthmaker and formal proofs do not allow an infinite sequence of
steps then we have an analytical truth with no proof yet it still has a
truthmaker.
That there cannot be any analytic truth without a truthmaker
refutes the Tarski Undecidability theorem.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof...(Gödel 1931:4344)
Tarski anchored his proof in an epistemological antinomy just like the
above quote: (3) x ∉ Provable if and only if x ∈ True.
https://liarparadox.org/Tarski_275_276.pdf
Epistemological antinomies cannot possibly have a truthmaker (not even
with an infinite number of steps) thus are simply untrue.
...14 Every epistemological antinomy can likewise be used for a similar undecidability proof...(Gödel 1931:4344)
Thus Gödel really screwed up. Epistemological antinomies are neither
true nor false thus calling them undecidable is a terrible euphemism
for nontruthbearer.
On 12/7/2023 11:29 PM, olcott wrote:
On 12/7/2023 6:11 PM, olcott wrote:
On 12/7/2023 10:20 AM, olcott wrote:
On 12/6/2023 9:56 PM, olcott wrote:
On 12/6/2023 4:35 PM, Jim Burns wrote:
On 12/6/2023 4:55 PM, olcott wrote:
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Fascinating.
Would you like me to tell you about
global warming?
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Consider reading the post to which you respond.
qnff =
 Q is notfirst false
finseq =
 If this finite sequence of claims
 holds a false claim,
 then it holds a first false claim.
Abbreviate
a definition of "n is a natural number"
as "n is a natrual number.
natnum =
 n is a natural number
 if and only if
 n satisfies the definition of natural number

Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
Here's the problem:
You:
That all seems to be gibberish to me.
For you to understand what I am saying you must learn a little
philosophy.
The Sapir–Whorf hypothesis shows that there may be some
concepts that cannot be expressed within the scope of the
terms of logic.
https://en.wikipedia.org/wiki/Linguistic_relativity
Everything that is true on the basis of its meaning:
https://en.wikipedia.org/wiki/Analytic%E2%80%93synthetic_distinction >>>>>
AKA the analytic side of the analytic / synthetic distinction
necessarily must have a connection from an expression
to this meaning as its truthmaker or it cannot possibly be true.
Although within the conventional terms of logic there
may be some truths that cannot be proven there cannot
be analytic expressions of language that are true without
something making them true.
I have diligently accounted for the difference between analytical truth >>>> and analytical knowledge the former may require an infinite sequence of >>>> steps as its truthmaker.
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
Eliminates Tarski undefinability and Gödel incompleteness and forces
the concept of truth in math and logic to conform to the way that it
works everywhere else in the body of human knowledge: True(x) ≡ (⊢ x) >>>>
If the Goldbach conjecture only has an infinite sequence of steps as
its truthmaker and formal proofs do not allow an infinite sequence of
steps then we have an analytical truth with no proof yet it still has a >>>> truthmaker.
That there cannot be any analytic truth without a truthmaker
refutes the Tarski Undecidability theorem.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof...(Gödel 1931:4344)
Tarski anchored his proof in an epistemological antinomy just like the
above quote: (3) x ∉ Provable if and only if x ∈ True.
https://liarparadox.org/Tarski_275_276.pdf
Epistemological antinomies cannot possibly have a truthmaker (not even
with an infinite number of steps) thus are simply untrue.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof...(Gödel 1931:4344)
Thus Gödel really screwed up. Epistemological antinomies are neither
true nor false thus calling them undecidable is a terrible euphemism
for nontruthbearer.
The last two paragraphs are to be analyzed in isolation.
Saying that Gödel did not screw up because other parts of
his paper did not screw up is the strawman deception.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof... (Gödel 1931:4344)
Thus Undecidable(L,x) is merely a terribly misleading euphemism for ~True(L,x).
On 12/8/2023 11:19 AM, olcott wrote:
On 12/7/2023 11:29 PM, olcott wrote:
On 12/7/2023 6:11 PM, olcott wrote:
On 12/7/2023 10:20 AM, olcott wrote:
On 12/6/2023 9:56 PM, olcott wrote:
On 12/6/2023 4:35 PM, Jim Burns wrote:
On 12/6/2023 4:55 PM, olcott wrote:
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Fascinating.
Would you like me to tell you about
global warming?
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Consider reading the post to which you respond.
qnff =
 Q is notfirst false
finseq =
 If this finite sequence of claims
 holds a false claim,
 then it holds a first false claim.
Abbreviate
a definition of "n is a natural number"
as "n is a natrual number.
natnum =
 n is a natural number
 if and only if
 n satisfies the definition of natural number

Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
Here's the problem:
You:
That all seems to be gibberish to me.
For you to understand what I am saying you must learn a little
philosophy.
The Sapir–Whorf hypothesis shows that there may be some
concepts that cannot be expressed within the scope of the
terms of logic.
https://en.wikipedia.org/wiki/Linguistic_relativity
Everything that is true on the basis of its meaning:
https://en.wikipedia.org/wiki/Analytic%E2%80%93synthetic_distinction >>>>>>
AKA the analytic side of the analytic / synthetic distinction
necessarily must have a connection from an expression
to this meaning as its truthmaker or it cannot possibly be true.
Although within the conventional terms of logic there
may be some truths that cannot be proven there cannot
be analytic expressions of language that are true without
something making them true.
I have diligently accounted for the difference between analytical
truth
and analytical knowledge the former may require an infinite
sequence of
steps as its truthmaker.
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
Eliminates Tarski undefinability and Gödel incompleteness and
forces the concept of truth in math and logic to conform to the way
that it works everywhere else in the body of human knowledge:
True(x) ≡ (⊢ x)
If the Goldbach conjecture only has an infinite sequence of steps as >>>>> its truthmaker and formal proofs do not allow an infinite sequence of >>>>> steps then we have an analytical truth with no proof yet it still
has a
truthmaker.
That there cannot be any analytic truth without a truthmaker
refutes the Tarski Undecidability theorem.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof...(Gödel 1931:4344)
Tarski anchored his proof in an epistemological antinomy just like the >>>> above quote: (3) x ∉ Provable if and only if x ∈ True.
https://liarparadox.org/Tarski_275_276.pdf
Epistemological antinomies cannot possibly have a truthmaker (not even >>>> with an infinite number of steps) thus are simply untrue.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof...(Gödel 1931:4344)
Thus Gödel really screwed up. Epistemological antinomies are neither
true nor false thus calling them undecidable is a terrible euphemism
for nontruthbearer.
The last two paragraphs are to be analyzed in isolation.
Saying that Gödel did not screw up because other parts of
his paper did not screw up is the strawman deception.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof... (Gödel 1931:4344)
Thus Undecidable(L,x) is merely a terribly misleading euphemism for
~True(L,x).
Undecidable has the base meaning that one cannot make up one's mind.
The mathematical use of the term includes the inability to decide
whether a kitten is a 15 story office building or a 16 story office
building and no option to say "incorrect question".
...14 Every epistemological antinomy can likewise be used
for a similar undecidability proof... (Gödel 1931:4344)
When a decision problem requires a yes or no answer to
an expression of language that is a selfcontradictory
question this is incorrectly ruled as undecidable.
On 12/8/2023 1:03 PM, olcott wrote:
On 12/8/2023 12:35 PM, olcott wrote:
On 12/8/2023 11:19 AM, olcott wrote:
On 12/7/2023 11:29 PM, olcott wrote:
On 12/7/2023 6:11 PM, olcott wrote:
On 12/7/2023 10:20 AM, olcott wrote:
On 12/6/2023 9:56 PM, olcott wrote:
On 12/6/2023 4:35 PM, Jim Burns wrote:
On 12/6/2023 4:55 PM, olcott wrote:
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Fascinating.
Would you like me to tell you about
global warming?
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Consider reading the post to which you respond.
qnff =
 Q is notfirst false
finseq =
 If this finite sequence of claims
 holds a false claim,
 then it holds a first false claim.
Abbreviate
a definition of "n is a natural number"
as "n is a natrual number.
natnum =
 n is a natural number
 if and only if
 n satisfies the definition of natural number

Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
Here's the problem:
You:
That all seems to be gibberish to me.
For you to understand what I am saying you must learn a little >>>>>>>> philosophy.
The Sapir–Whorf hypothesis shows that there may be some
concepts that cannot be expressed within the scope of the
terms of logic.
https://en.wikipedia.org/wiki/Linguistic_relativity
Everything that is true on the basis of its meaning:
https://en.wikipedia.org/wiki/Analytic%E2%80%93synthetic_distinction >>>>>>>>
AKA the analytic side of the analytic / synthetic distinction
necessarily must have a connection from an expression
to this meaning as its truthmaker or it cannot possibly be true. >>>>>>>>
Although within the conventional terms of logic there
may be some truths that cannot be proven there cannot
be analytic expressions of language that are true without
something making them true.
I have diligently accounted for the difference between analytical >>>>>>> truth
and analytical knowledge the former may require an infinite
sequence of
steps as its truthmaker.
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
Eliminates Tarski undefinability and Gödel incompleteness and
forces the concept of truth in math and logic to conform to the
way that it works everywhere else in the body of human knowledge: >>>>>>> True(x) ≡ (⊢ x)
If the Goldbach conjecture only has an infinite sequence of steps as >>>>>>> its truthmaker and formal proofs do not allow an infinite
sequence of
steps then we have an analytical truth with no proof yet it still >>>>>>> has a
truthmaker.
That there cannot be any analytic truth without a truthmaker
refutes the Tarski Undecidability theorem.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof...(Gödel 1931:4344)
Tarski anchored his proof in an epistemological antinomy just like >>>>>> the
above quote: (3) x ∉ Provable if and only if x ∈ True.
https://liarparadox.org/Tarski_275_276.pdf
Epistemological antinomies cannot possibly have a truthmaker (not
even
with an infinite number of steps) thus are simply untrue.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof...(Gödel 1931:4344)
Thus Gödel really screwed up. Epistemological antinomies are neither >>>>> true nor false thus calling them undecidable is a terrible euphemism >>>>> for nontruthbearer.
The last two paragraphs are to be analyzed in isolation.
Saying that Gödel did not screw up because other parts of
his paper did not screw up is the strawman deception.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof... (Gödel 1931:4344)
Thus Undecidable(L,x) is merely a terribly misleading euphemism for
~True(L,x).
Undecidable has the base meaning that one cannot make up one's mind.
The mathematical use of the term includes the inability to decide
whether a kitten is a 15 story office building or a 16 story office
building and no option to say "incorrect question".
...14 Every epistemological antinomy can likewise be used
for a similar undecidability proof... (Gödel 1931:4344)
When a decision problem requires a yes or no answer to
an expression of language that is a selfcontradictory
question this is incorrectly ruled as undecidable.
"As I asked before, show exactly WHERE he is doing what you claim."
As I have said 500 times! RIGHT HERE !!!
...14 Every epistemological antinomy can likewise be used
for a similar undecidability proof... (Gödel 1931:4344)
The truth is that NO epistemological antinomies CAN EVER
be used in ANY undecidability proof, not ever not even once.
On 12/8/2023 1:37 PM, olcott wrote:
On 12/8/2023 1:03 PM, olcott wrote:
On 12/8/2023 12:35 PM, olcott wrote:
On 12/8/2023 11:19 AM, olcott wrote:
On 12/7/2023 11:29 PM, olcott wrote:
On 12/7/2023 6:11 PM, olcott wrote:
On 12/7/2023 10:20 AM, olcott wrote:
On 12/6/2023 9:56 PM, olcott wrote:
On 12/6/2023 4:35 PM, Jim Burns wrote:
On 12/6/2023 4:55 PM, olcott wrote:
On 12/6/2023 3:49 PM, Jim Burns wrote:
On 12/6/2023 4:37 PM, olcott wrote:
It seems that you ignored all ofis
my important points.
excludes unknown truths
your most important point.
YMMV.
Unknown truths are at least as capable of
killing you as known truths.
Philosophically I am only referring to
the analytic side of
the analytic / synthetic distinction.
That excludes physical reality where
things can kill you.
Fascinating.
Would you like me to tell you about
global warming?
Can you try again and ask to have
anything that you don't understand
explained?
True(qnff) ?
¬True(qnff) ?
True(finseq) ?
¬True(finseq) ?
True(natnum) ?
¬True(natnum) ?
That all seems to be gibberish to me.
Consider reading the post to which you respond.
qnff =
 Q is notfirst false
finseq =
 If this finite sequence of claims
 holds a false claim,
 then it holds a first false claim.
Abbreviate
a definition of "n is a natural number"
as "n is a natrual number.
natnum =
 n is a natural number
 if and only if
 n satisfies the definition of natural number

Consider the claim qnff =
 Q is notfirstfalse in
 ⟨… P∨Q ¬P Q …⟩
 t f t
 t t t
 t f f
 f t f

True(qnff) ?
¬True(qnff) ?
Consider the claim finseq =
 For finite sequence ⟨foo … bar⟩
 if ⟨foo … bar⟩ holds a false claim,
 then it holds a first false claim.

True(finseq) ?
¬True(finseq) ?
_Abbreviate_
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1
as
 n is a natural number
Consider the claim natnum =
 n is a natural number
 if and only if
 n ends ordered ⟨0,…,n⟩ such that,
 for each split Fᣔ<ᣔH of ⟨0,…,n⟩
 some i‖i+1 is last‖first in F‖H, and
 0‖n is first‖last in ⟨0,…,n⟩
 for
 non0 nondoppelgänger nonfinal i+1

True(natnum) ?
¬True(natnum) ?
Are you expecting these answers to change
if, for example, a proof of the Goldbach
conjecture is discovered?
Please explain.
Do you understand what this steps of
the Tarski proof says:
(3) x ∉ Provable if and only if x ∈ True.
Here's the problem:
You:
That all seems to be gibberish to me.
For you to understand what I am saying you must learn a little >>>>>>>>> philosophy.
The Sapir–Whorf hypothesis shows that there may be some
concepts that cannot be expressed within the scope of the
terms of logic.
https://en.wikipedia.org/wiki/Linguistic_relativity
Everything that is true on the basis of its meaning:
https://en.wikipedia.org/wiki/Analytic%E2%80%93synthetic_distinction >>>>>>>>>
AKA the analytic side of the analytic / synthetic distinction >>>>>>>>> necessarily must have a connection from an expression
to this meaning as its truthmaker or it cannot possibly be true. >>>>>>>>>
Although within the conventional terms of logic there
may be some truths that cannot be proven there cannot
be analytic expressions of language that are true without
something making them true.
I have diligently accounted for the difference between
analytical truth
and analytical knowledge the former may require an infinite
sequence of
steps as its truthmaker.
∀L ∈ Formal_System ∀x ∈ Language(L)
True(L,x) ≡ (T ⊢ x)
False(L,x) ≡ (T ⊢ ¬x)
Eliminates Tarski undefinability and Gödel incompleteness and >>>>>>>> forces the concept of truth in math and logic to conform to the >>>>>>>> way that it works everywhere else in the body of human
knowledge: True(x) ≡ (⊢ x)
If the Goldbach conjecture only has an infinite sequence of
steps as
its truthmaker and formal proofs do not allow an infinite
sequence of
steps then we have an analytical truth with no proof yet it
still has a
truthmaker.
That there cannot be any analytic truth without a truthmaker
refutes the Tarski Undecidability theorem.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof...(Gödel 1931:4344)
Tarski anchored his proof in an epistemological antinomy just
like the
above quote: (3) x ∉ Provable if and only if x ∈ True.
https://liarparadox.org/Tarski_275_276.pdf
Epistemological antinomies cannot possibly have a truthmaker (not >>>>>>> even
with an infinite number of steps) thus are simply untrue.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof...(Gödel 1931:4344)
Thus Gödel really screwed up. Epistemological antinomies are neither >>>>>> true nor false thus calling them undecidable is a terrible euphemism >>>>>> for nontruthbearer.
The last two paragraphs are to be analyzed in isolation.
Saying that Gödel did not screw up because other parts of
his paper did not screw up is the strawman deception.
...14 Every epistemological antinomy can likewise be used for a
similar undecidability proof... (Gödel 1931:4344)
Thus Undecidable(L,x) is merely a terribly misleading euphemism for
~True(L,x).
Undecidable has the base meaning that one cannot make up one's mind.
The mathematical use of the term includes the inability to decide
whether a kitten is a 15 story office building or a 16 story office
building and no option to say "incorrect question".
...14 Every epistemological antinomy can likewise be used
for a similar undecidability proof... (Gödel 1931:4344)
When a decision problem requires a yes or no answer to
an expression of language that is a selfcontradictory
question this is incorrectly ruled as undecidable.
"As I asked before, show exactly WHERE he is doing what you claim."
As I have said 500 times! RIGHT HERE !!!
...14 Every epistemological antinomy can likewise be used
for a similar undecidability proof... (Gödel 1931:4344)
The truth is that NO epistemological antinomies CAN EVER
be used in ANY undecidability proof, not ever not even once.
That Gödel would say this proves that he did not have a clue
about the subject matter of his paper.
On 12/7/2023 6:11 PM, olcott wrote:
On 12/7/2023 10:20 AM, olcott wrote:
On 12/6/2023 9:56 PM, olcott wrote:
On 12/6/2023 4:35 PM, Jim Burns wrote:
[...]
...14
Every epistemological antinomy can likewise
be used for a similar undecidability proof...
(Gödel 1931:4344)
Thus Gödel really screwed up.
Epistemological antinomies
Epistemological antinomies
are neither
On 12/9/2023 7:22 PM, Jim Burns wrote:
On 12/8/2023 12:29 AM, olcott wrote:
On 12/7/2023 6:11 PM, olcott wrote:
On 12/7/2023 10:20 AM, olcott wrote:
On 12/6/2023 9:56 PM, olcott wrote:
On 12/6/2023 4:35 PM, Jim Burns wrote:
[...]
...14
Every epistemological antinomy can likewise
be used for a similar undecidability proof...
(Gödel 1931:4344)
Thus Gödel really screwed up.
Epistemological antinomies
The epistemological antinomy
 This sentence is false

is not in Gödel's proof.
 This sentence is false

is the blueprint, which guides
Gödel placement of (metaphorically) actual
bricks and mortar.
You live in a building of some kind, I'd bet.
What odds would you give on whether
that building's blueprints are incorporated
into its construction?
If you ripped plaster off walls,
would you find particular sheets paper
holding up waterlines?
In note 14, Gödel is mentioning that
other blueprints can guide the placement of
(metaphorically) actual bricks and mortar
for other proofs.
Nor are those other blueprints incorporated
into those other proofs.
Epistemological antinomies
are neither
...here nor there.
Since no epistemological antinomy can ever be used for
any proof at all Gödel proved that it didn't have a clue
about the subject matter of his paper.
https://liarparadox.org/Tarski_247_248.pdf
Tarski said that he used Gödel as a basis
and in the above link shows that he anchored
his whole proof in the actual Liar Paradox.
*Here is his actual proof*
https://liarparadox.org/Tarski_275_276.pdf
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