XPost: sci.logic

Op 05.feb.2024 om 05:19 schreef olcott:

On 2/4/2024 9:37 PM, Richard Damon wrote:

On 2/4/24 10:14 PM, olcott wrote:

On 2/4/2024 8:29 PM, Richard Damon wrote:

On 2/4/24 8:44 PM, olcott wrote:

On 2/4/2024 5:49 PM, Richard Damon wrote:

On 2/4/24 4:14 PM, olcott wrote:

On 2/4/2024 12:16 PM, Richard Damon wrote:

On 2/4/24 1:02 PM, olcott wrote:

On 2/4/2024 11:45 AM, Richard Damon wrote:

On 2/4/24 10:53 AM, olcott wrote:

An analytic expression x is any expression of language
verified as
completely true (or false) entirely on the basis that x (or >>>>>>>>>>> ~x) is
derived by applying truth preserving operations to other >>>>>>>>>>> expressions
of language that are stipulated to be true thus providing the >>>>>>>>>>> semantic
meaning of terms.

Right

...14 Every epistemological antinomy can likewise be used for >>>>>>>>>>> a similar undecidability proof...(Gödel 1931:43)

WHich you just don't understand what they did.

When we understand that a True(L, x) predicate only applies >>>>>>>>>>> to analytic
expressions then Tarski's proof utterly fails because
epistemological antinomies are rejected as not analytic.

So, the fact that the EXISTANCE of a computabe True(L, x), >>>>>>>>>> allows the proving of that an epistemological antinomy must >>>>>>>>>> say that such a thing can not exist doesn't matter to you? >>>>>>>>>>

A truth predicate only deals with analytic expressions, everything >>>>>>>>> else is out-of-scope. A truth predicate rejects expressions that >>>>>>>>> are not analytic.

A Truth Predicate needs to deal with ALL expressions in the
Language.

IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies

It must say they are not true, (and not false).

That would also work, yet we construe not true and not false
as not truth bearer thus out-of-scope of a truth predicate.

But they are answer of different predicates.

We only need to Truth predicate and this must reject
epistemological antinomies. Neither Tarski nor Gödel
understood this.

*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*

But you don't understand what he is doing,

Just like a Halt Decider must handle all programs definable in >>>>>>>> the system.

This is not the same because D correctly simulated by H cannot
possibly
reach the self-contradiction.

No, because the only H that "Correctly Simulates" its input is the >>>>>> one that never aborts it.

You continue to fail to understand that correctly simulating N steps >>>>> <IS> a correct simulation of these N steps.

But NOT a "Correct Simulation" that allows the use of a the
simulation to replace the behavior of the machine described.

Both Hehner and Stoddart agree that:
D does specify non terminating behavior to H.

*Maybe these are finally the right words*
D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

Lets call the H that aborts after N steps HaN.
For each of these we can create a DaN based on this HaN.

It *is* the case that a halt decider must compute
the mapping from the behavior that its finite string
specifies...

HaN(DaN,DaN) must judge in input, which contains HaN, which aborts after
N steps and therefore DaN halts. It should not judge its non-input the
DaM, which aborts after M steps. (With M ≠ N.) There is no N for which
HaH gets a correct result for HaN(DaN,DaN).

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: sci.logic

On 2/4/24 11:19 PM, olcott wrote:

On 2/4/2024 9:37 PM, Richard Damon wrote:

On 2/4/24 10:14 PM, olcott wrote:

On 2/4/2024 8:29 PM, Richard Damon wrote:

On 2/4/24 8:44 PM, olcott wrote:

On 2/4/2024 5:49 PM, Richard Damon wrote:

On 2/4/24 4:14 PM, olcott wrote:

On 2/4/2024 12:16 PM, Richard Damon wrote:

On 2/4/24 1:02 PM, olcott wrote:

On 2/4/2024 11:45 AM, Richard Damon wrote:

On 2/4/24 10:53 AM, olcott wrote:

An analytic expression x is any expression of language
verified as
completely true (or false) entirely on the basis that x (or >>>>>>>>>>> ~x) is
derived by applying truth preserving operations to other >>>>>>>>>>> expressions
of language that are stipulated to be true thus providing the >>>>>>>>>>> semantic
meaning of terms.

Right

...14 Every epistemological antinomy can likewise be used for >>>>>>>>>>> a similar undecidability proof...(Gödel 1931:43)

WHich you just don't understand what they did.

When we understand that a True(L, x) predicate only applies >>>>>>>>>>> to analytic
expressions then Tarski's proof utterly fails because
epistemological antinomies are rejected as not analytic.

So, the fact that the EXISTANCE of a computabe True(L, x), >>>>>>>>>> allows the proving of that an epistemological antinomy must >>>>>>>>>> say that such a thing can not exist doesn't matter to you? >>>>>>>>>>

A truth predicate only deals with analytic expressions, everything >>>>>>>>> else is out-of-scope. A truth predicate rejects expressions that >>>>>>>>> are not analytic.

A Truth Predicate needs to deal with ALL expressions in the
Language.

IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies

It must say they are not true, (and not false).

That would also work, yet we construe not true and not false
as not truth bearer thus out-of-scope of a truth predicate.

But they are answer of different predicates.

We only need to Truth predicate and this must reject
epistemological antinomies. Neither Tarski nor Gödel
understood this.

*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*

But you don't understand what he is doing,

Just like a Halt Decider must handle all programs definable in >>>>>>>> the system.

This is not the same because D correctly simulated by H cannot
possibly
reach the self-contradiction.

No, because the only H that "Correctly Simulates" its input is the >>>>>> one that never aborts it.

You continue to fail to understand that correctly simulating N steps >>>>> <IS> a correct simulation of these N steps.

But NOT a "Correct Simulation" that allows the use of a the
simulation to replace the behavior of the machine described.

Both Hehner and Stoddart agree that:
D does specify non terminating behavior to H.

*Maybe these are finally the right words*
D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

It *is* the case that a halt decider must compute
the mapping from the behavior that its finite string
specifies...

So, you are just agreeing that your H is just a POOP decider, not a Halt Decider.

If your criteria is not the actual Halting Question, your results aren't
a Halt Decider,

For a Halt Decider, it must compute the mapping from the finite input
string to the behavior that string specifies, and that string specifies
a Compuation, and the FULL sequnece of states that running it will
generate, and we ask if the sequence has an end.

It is NOT about can H simulate that input to the end, but does it have one.

And, for this case, the input is built on a copy of the EXACT
computation that is deciding on it.

You don't get to change the criteria to something that creates a
different map.

You clearly don't understand the nature of requirements, and thus the
meaning of "Correct" or "True"

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: sci.logic

Op 05.feb.2024 om 15:28 schreef olcott:

On 2/5/2024 3:52 AM, Fred. Zwarts wrote:

Op 05.feb.2024 om 05:19 schreef olcott:

On 2/4/2024 9:37 PM, Richard Damon wrote:

On 2/4/24 10:14 PM, olcott wrote:

On 2/4/2024 8:29 PM, Richard Damon wrote:

On 2/4/24 8:44 PM, olcott wrote:

On 2/4/2024 5:49 PM, Richard Damon wrote:

On 2/4/24 4:14 PM, olcott wrote:

On 2/4/2024 12:16 PM, Richard Damon wrote:

On 2/4/24 1:02 PM, olcott wrote:

On 2/4/2024 11:45 AM, Richard Damon wrote:

On 2/4/24 10:53 AM, olcott wrote:

An analytic expression x is any expression of language >>>>>>>>>>>>> verified as
completely true (or false) entirely on the basis that x (or >>>>>>>>>>>>> ~x) is
derived by applying truth preserving operations to other >>>>>>>>>>>>> expressions
of language that are stipulated to be true thus providing >>>>>>>>>>>>> the semantic
meaning of terms.

Right

...14 Every epistemological antinomy can likewise be used >>>>>>>>>>>>> for a similar undecidability proof...(Gödel 1931:43) >>>>>>>>>>>>>

WHich you just don't understand what they did.

When we understand that a True(L, x) predicate only applies >>>>>>>>>>>>> to analytic
expressions then Tarski's proof utterly fails because >>>>>>>>>>>>> epistemological antinomies are rejected as not analytic. >>>>>>>>>>>>

So, the fact that the EXISTANCE of a computabe True(L, x), >>>>>>>>>>>> allows the proving of that an epistemological antinomy must >>>>>>>>>>>> say that such a thing can not exist doesn't matter to you? >>>>>>>>>>>>

A truth predicate only deals with analytic expressions,
everything
else is out-of-scope. A truth predicate rejects expressions that >>>>>>>>>>> are not analytic.

A Truth Predicate needs to deal with ALL expressions in the >>>>>>>>>> Language.

IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies

It must say they are not true, (and not false).

That would also work, yet we construe not true and not false
as not truth bearer thus out-of-scope of a truth predicate.

But they are answer of different predicates.

We only need to Truth predicate and this must reject
epistemological antinomies. Neither Tarski nor Gödel
understood this.

*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*

But you don't understand what he is doing,

Just like a Halt Decider must handle all programs definable in >>>>>>>>>> the system.

This is not the same because D correctly simulated by H cannot >>>>>>>>> possibly
reach the self-contradiction.

No, because the only H that "Correctly Simulates" its input is >>>>>>>> the one that never aborts it.

You continue to fail to understand that correctly simulating N steps >>>>>>> <IS> a correct simulation of these N steps.

But NOT a "Correct Simulation" that allows the use of a the
simulation to replace the behavior of the machine described.

Both Hehner and Stoddart agree that:
D does specify non terminating behavior to H.

*Maybe these are finally the right words*
D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

Lets call the H that aborts after N steps HaN.
For each of these we can create a DaN based on this HaN.

When H correctly reports on the behavior of every H that can
possibly exist we need no other H.

It *is* the case that a halt decider must compute
the mapping from the behavior that its finite string
specifies...

HaN(DaN,DaN) must judge in input, which contains HaN, which aborts
after N steps and therefore DaN halts. It should not judge its
non-input the DaM, which aborts after M steps. (With M ≠ N.) There is
no N for which HaH gets a correct result for HaN(DaN,DaN).

D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

Each of the DaN specify that it reaches the final state, so, if H
aborts, there is no D that does not reach the final state.

As soon as H correctly determines that the above is
true in N steps of correct simulation then H correctly
aborts its simulated and rejects D as non-halting.

None of the HaN correctly simulate DaN, because they all need to
simulate a few steps more than they do. A H that simulates N steps and
more than N steps at the same time does not exist. So, there is no H
that correctly determines that the above is true in N steps, because it
always needs more steps than N. If H simulates N steps, it needs a few
steps more to see that DaN returns. No correct HaN exists, not even in
the limit N → ∞.
Remember That HaN(DaN,DaN) must decide on its input DaN, not on its
non-input DaM, with M ≠ N.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: sci.logic

Op 05.feb.2024 om 16:15 schreef olcott:

On 2/5/2024 8:58 AM, Fred. Zwarts wrote:

Op 05.feb.2024 om 15:28 schreef olcott:

On 2/5/2024 3:52 AM, Fred. Zwarts wrote:

Op 05.feb.2024 om 05:19 schreef olcott:

On 2/4/2024 9:37 PM, Richard Damon wrote:

On 2/4/24 10:14 PM, olcott wrote:

On 2/4/2024 8:29 PM, Richard Damon wrote:

On 2/4/24 8:44 PM, olcott wrote:

On 2/4/2024 5:49 PM, Richard Damon wrote:

On 2/4/24 4:14 PM, olcott wrote:

On 2/4/2024 12:16 PM, Richard Damon wrote:

On 2/4/24 1:02 PM, olcott wrote:

On 2/4/2024 11:45 AM, Richard Damon wrote:

On 2/4/24 10:53 AM, olcott wrote:

An analytic expression x is any expression of language >>>>>>>>>>>>>>> verified as
completely true (or false) entirely on the basis that x >>>>>>>>>>>>>>> (or ~x) is
derived by applying truth preserving operations to other >>>>>>>>>>>>>>> expressions
of language that are stipulated to be true thus providing >>>>>>>>>>>>>>> the semantic
meaning of terms.

Right

...14 Every epistemological antinomy can likewise be used >>>>>>>>>>>>>>> for a similar undecidability proof...(Gödel 1931:43) >>>>>>>>>>>>>>>

WHich you just don't understand what they did.

When we understand that a True(L, x) predicate only >>>>>>>>>>>>>>> applies to analytic
expressions then Tarski's proof utterly fails because >>>>>>>>>>>>>>> epistemological antinomies are rejected as not analytic. >>>>>>>>>>>>>>

So, the fact that the EXISTANCE of a computabe True(L, x), >>>>>>>>>>>>>> allows the proving of that an epistemological antinomy >>>>>>>>>>>>>> must say that such a thing can not exist doesn't matter to >>>>>>>>>>>>>> you?

A truth predicate only deals with analytic expressions, >>>>>>>>>>>>> everything
else is out-of-scope. A truth predicate rejects expressions >>>>>>>>>>>>> that
are not analytic.

A Truth Predicate needs to deal with ALL expressions in the >>>>>>>>>>>> Language.

IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies

It must say they are not true, (and not false).

That would also work, yet we construe not true and not false >>>>>>>>> as not truth bearer thus out-of-scope of a truth predicate.

But they are answer of different predicates.

We only need to Truth predicate and this must reject
epistemological antinomies. Neither Tarski nor Gödel
understood this.

*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*

But you don't understand what he is doing,

Just like a Halt Decider must handle all programs definable >>>>>>>>>>>> in the system.

This is not the same because D correctly simulated by H
cannot possibly
reach the self-contradiction.

No, because the only H that "Correctly Simulates" its input is >>>>>>>>>> the one that never aborts it.

You continue to fail to understand that correctly simulating N >>>>>>>>> steps
<IS> a correct simulation of these N steps.

But NOT a "Correct Simulation" that allows the use of a the
simulation to replace the behavior of the machine described.

Both Hehner and Stoddart agree that:
D does specify non terminating behavior to H.

*Maybe these are finally the right words*
D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

Lets call the H that aborts after N steps HaN.
For each of these we can create a DaN based on this HaN.

When H correctly reports on the behavior of every H that can
possibly exist we need no other H.

It *is* the case that a halt decider must compute
the mapping from the behavior that its finite string
specifies...

HaN(DaN,DaN) must judge in input, which contains HaN, which aborts
after N steps and therefore DaN halts. It should not judge its
non-input the DaM, which aborts after M steps. (With M ≠ N.) There
is no N for which HaH gets a correct result for HaN(DaN,DaN).

D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

Each of the DaN specify that it reaches the final state, so, if H
aborts, there is no D that does not reach the final state.

As soon as H correctly determines that the above is
true in N steps of correct simulation then H correctly
aborts its simulated and rejects D as non-halting.

None of the HaN correctly simulate DaN, because they all need to

Of the infinite set of every H1...Hn that correctly simulates
1 to ∞ steps of its corresponding D1...Dn none of these infinite
pairs ever reaches its own final state.

That should be:
Of the infinite set of every H1...Hn that simulates 1 to n steps, none simulates enough steps to see that D1 halts normally. So, they all abort
too soon and return falsely an non-halting state.

The specific H that correctly simulates N steps of D correctly
determines that above is true (for the entire infinite set)
thus providing it with the correct halt status criteria basis
to reject D as non-halting.

That should be:
The fact that a true simulator that simulates Dn with enough steps halts normally, proves that this is true for the entire infinite set, thus
proving that none of these H1...Hn correctly report a non-halting state.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: sci.logic

Op 05.feb.2024 om 16:43 schreef Fred. Zwarts:

Op 05.feb.2024 om 16:15 schreef olcott:

On 2/5/2024 8:58 AM, Fred. Zwarts wrote:

Op 05.feb.2024 om 15:28 schreef olcott:

On 2/5/2024 3:52 AM, Fred. Zwarts wrote:

Op 05.feb.2024 om 05:19 schreef olcott:

On 2/4/2024 9:37 PM, Richard Damon wrote:

On 2/4/24 10:14 PM, olcott wrote:

On 2/4/2024 8:29 PM, Richard Damon wrote:

On 2/4/24 8:44 PM, olcott wrote:

On 2/4/2024 5:49 PM, Richard Damon wrote:

On 2/4/24 4:14 PM, olcott wrote:

On 2/4/2024 12:16 PM, Richard Damon wrote:

On 2/4/24 1:02 PM, olcott wrote:

On 2/4/2024 11:45 AM, Richard Damon wrote:

On 2/4/24 10:53 AM, olcott wrote:

An analytic expression x is any expression of language >>>>>>>>>>>>>>>> verified as
completely true (or false) entirely on the basis that x >>>>>>>>>>>>>>>> (or ~x) is
derived by applying truth preserving operations to other >>>>>>>>>>>>>>>> expressions
of language that are stipulated to be true thus >>>>>>>>>>>>>>>> providing the semantic
meaning of terms.

Right

...14 Every epistemological antinomy can likewise be >>>>>>>>>>>>>>>> used for a similar undecidability proof...(Gödel 1931:43) >>>>>>>>>>>>>>>>

WHich you just don't understand what they did.

When we understand that a True(L, x) predicate only >>>>>>>>>>>>>>>> applies to analytic
expressions then Tarski's proof utterly fails because >>>>>>>>>>>>>>>> epistemological antinomies are rejected as not analytic. >>>>>>>>>>>>>>>

So, the fact that the EXISTANCE of a computabe True(L, >>>>>>>>>>>>>>> x), allows the proving of that an epistemological >>>>>>>>>>>>>>> antinomy must say that such a thing can not exist doesn't >>>>>>>>>>>>>>> matter to you?

A truth predicate only deals with analytic expressions, >>>>>>>>>>>>>> everything
else is out-of-scope. A truth predicate rejects
expressions that
are not analytic.

A Truth Predicate needs to deal with ALL expressions in the >>>>>>>>>>>>> Language.

IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies

It must say they are not true, (and not false).

That would also work, yet we construe not true and not false >>>>>>>>>> as not truth bearer thus out-of-scope of a truth predicate. >>>>>>>>>

But they are answer of different predicates.

We only need to Truth predicate and this must reject
epistemological antinomies. Neither Tarski nor Gödel
understood this.

*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*

But you don't understand what he is doing,

Just like a Halt Decider must handle all programs definable >>>>>>>>>>>>> in the system.

This is not the same because D correctly simulated by H >>>>>>>>>>>> cannot possibly
reach the self-contradiction.

No, because the only H that "Correctly Simulates" its input >>>>>>>>>>> is the one that never aborts it.

You continue to fail to understand that correctly simulating N >>>>>>>>>> steps
<IS> a correct simulation of these N steps.

But NOT a "Correct Simulation" that allows the use of a the
simulation to replace the behavior of the machine described. >>>>>>>>>

Both Hehner and Stoddart agree that:
D does specify non terminating behavior to H.

*Maybe these are finally the right words*
D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

Lets call the H that aborts after N steps HaN.
For each of these we can create a DaN based on this HaN.

When H correctly reports on the behavior of every H that can
possibly exist we need no other H.

It *is* the case that a halt decider must compute
the mapping from the behavior that its finite string
specifies...

HaN(DaN,DaN) must judge in input, which contains HaN, which aborts
after N steps and therefore DaN halts. It should not judge its
non-input the DaM, which aborts after M steps. (With M ≠ N.) There >>>>> is no N for which HaH gets a correct result for HaN(DaN,DaN).

D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

Each of the DaN specify that it reaches the final state, so, if H
aborts, there is no D that does not reach the final state.

As soon as H correctly determines that the above is
true in N steps of correct simulation then H correctly
aborts its simulated and rejects D as non-halting.

None of the HaN correctly simulate DaN, because they all need to

Of the infinite set of every H1...Hn that correctly simulates
1 to ∞ steps of its corresponding D1...Dn none of these infinite
pairs ever reaches its own final state.

That should be:
Of the infinite set of every H1...Hn that simulates 1 to n steps, none simulates enough steps to see that D1 halts normally. So, they all abort
too soon and return falsely an non-halting state.

D1 should be Dn.

The specific H that correctly simulates N steps of D correctly
determines that above is true (for the entire infinite set)
thus providing it with the correct halt status criteria basis
to reject D as non-halting.

That should be:
The fact that a true simulator that simulates Dn with enough steps halts normally, proves that this is true for the entire infinite set, thus
proving that none of these H1...Hn correctly report a non-halting state.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: sci.logic

On 5/02/24 05:19, olcott wrote:

*Maybe these are finally the right words*
D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

It is impossible to simulate ∞ steps.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: sci.logic

On 5/02/24 19:21, olcott wrote:

On 2/5/2024 9:43 AM, Fred. Zwarts wrote:

Op 05.feb.2024 om 16:15 schreef olcott:

On 2/5/2024 8:58 AM, Fred. Zwarts wrote:

None of the HaN correctly simulate DaN, because they all need to

Of the infinite set of every H1...Hn that correctly simulates
1 to ∞ steps of its corresponding D1...Dn none of these infinite
pairs ever reaches its own final state.

That should be:
Of the infinite set of every H1...Hn that simulates 1 to n steps, none
simulates enough steps to see that D1 halts normally. So, they all
abort too soon and return falsely an non-halting state.

According to your reasoning no one has any idea that
Infinite_Loop() is an infinite loop until they run
it and wait until the end of time to find out that
it never halts.

void Infinite_Loop()
{
  HERE: goto HERE;
}

It is impossible for a simulating termination analyser to analyse
termination by any method other than simulation. If a termination
analyser uses something other than simulation, it is not as simulating termination analyser.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: sci.logic

On 5/02/24 15:18, olcott wrote:

D specifies that N to ∞ steps of D *correctly* simulated
by H cannot possibly reach the final state of D.

This doesn't make sense. You can't simulate ∞ steps.

Innovation DOES CHANGE THINGS.
No one ever thought of a simulating halt decider before.

Simulation was literally everyone's first thought when they thought
about how to make a halt decider. They thought some more, and realized
that it doesn't work and can't work.

Everyone that thought of simulation rejected it as a basis.

Because it doesn't work.

D specifies that N to ∞ steps of D *correctly* simulated
by H cannot possibly reach the final state of D.

This doesn't make sense. You can't simulate ∞ steps.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: sci.logic

On 2/5/24 9:28 AM, olcott wrote:

On 2/5/2024 3:52 AM, Fred. Zwarts wrote:

Op 05.feb.2024 om 05:19 schreef olcott:

On 2/4/2024 9:37 PM, Richard Damon wrote:

On 2/4/24 10:14 PM, olcott wrote:

On 2/4/2024 8:29 PM, Richard Damon wrote:

On 2/4/24 8:44 PM, olcott wrote:

On 2/4/2024 5:49 PM, Richard Damon wrote:

On 2/4/24 4:14 PM, olcott wrote:

On 2/4/2024 12:16 PM, Richard Damon wrote:

On 2/4/24 1:02 PM, olcott wrote:

On 2/4/2024 11:45 AM, Richard Damon wrote:

On 2/4/24 10:53 AM, olcott wrote:

An analytic expression x is any expression of language >>>>>>>>>>>>> verified as
completely true (or false) entirely on the basis that x (or >>>>>>>>>>>>> ~x) is
derived by applying truth preserving operations to other >>>>>>>>>>>>> expressions
of language that are stipulated to be true thus providing >>>>>>>>>>>>> the semantic
meaning of terms.

Right

...14 Every epistemological antinomy can likewise be used >>>>>>>>>>>>> for a similar undecidability proof...(Gödel 1931:43) >>>>>>>>>>>>>

WHich you just don't understand what they did.

When we understand that a True(L, x) predicate only applies >>>>>>>>>>>>> to analytic
expressions then Tarski's proof utterly fails because >>>>>>>>>>>>> epistemological antinomies are rejected as not analytic. >>>>>>>>>>>>

So, the fact that the EXISTANCE of a computabe True(L, x), >>>>>>>>>>>> allows the proving of that an epistemological antinomy must >>>>>>>>>>>> say that such a thing can not exist doesn't matter to you? >>>>>>>>>>>>

A truth predicate only deals with analytic expressions,
everything
else is out-of-scope. A truth predicate rejects expressions that >>>>>>>>>>> are not analytic.

A Truth Predicate needs to deal with ALL expressions in the >>>>>>>>>> Language.

IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies
IT MUST REJECT epistemological antinomies

It must say they are not true, (and not false).

That would also work, yet we construe not true and not false
as not truth bearer thus out-of-scope of a truth predicate.

But they are answer of different predicates.

We only need to Truth predicate and this must reject
epistemological antinomies. Neither Tarski nor Gödel
understood this.

*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*
*That is the key essence of their huge mistake*

But you don't understand what he is doing,

Just like a Halt Decider must handle all programs definable in >>>>>>>>>> the system.

This is not the same because D correctly simulated by H cannot >>>>>>>>> possibly
reach the self-contradiction.

No, because the only H that "Correctly Simulates" its input is >>>>>>>> the one that never aborts it.

You continue to fail to understand that correctly simulating N steps >>>>>>> <IS> a correct simulation of these N steps.

But NOT a "Correct Simulation" that allows the use of a the
simulation to replace the behavior of the machine described.

Both Hehner and Stoddart agree that:
D does specify non terminating behavior to H.

*Maybe these are finally the right words*
D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

Lets call the H that aborts after N steps HaN.
For each of these we can create a DaN based on this HaN.

When H correctly reports on the behavior of every H that can
possibly exist we need no other H.

And when if incorrectly reports on the actual behavior of its input, it
is just wrong as a Halt decider.

You are just proving you have POOP on your brain.

It *is* the case that a halt decider must compute
the mapping from the behavior that its finite string
specifies...

HaN(DaN,DaN) must judge in input, which contains HaN, which aborts
after N steps and therefore DaN halts. It should not judge its
non-input the DaM, which aborts after M steps. (With M ≠ N.) There is
no N for which HaH gets a correct result for HaN(DaN,DaN).

D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

But once you fix the H to a given number of steps, then

As soon as H correctly determines that the above is
true in N steps of correct simulation then H correctly
aborts its simulated and rejects D as non-halting.

So H is just a correct POOP decider, but not a Halt Decider, since a
Halt Decider needs to answer about the behavior of the specific input
given to it, and for any H that aborts this input and returns
non-halting, that input, which was based on THAT specific H, will halt
when run, and no H in your set every correctly simulated that specific
input (as the only one that was given it, incorrectly gave up too soon).

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: sci.logic

On 2/5/24 2:32 PM, immibis wrote:

On 5/02/24 05:19, olcott wrote:

*Maybe these are finally the right words*
D specifies that N to ∞ steps of D correctly simulated
by H cannot possibly reach the final state of D.

It is impossible to simulate ∞ steps.

No, depending on your exact definitions, you can simulate an infinite
number of steps, just not in finite time, which means the simulator can
not be a decider.

A UTM given the description of a Non-Halting compuation will simulate it
for an infinite number of steps.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)