On 10/23/2026 3:54 AM, Python wrote:

*keeps repeating unless aborted*
Line 03: simulated DD() invokes simulated HHH(DD) that simulates DD()

*Simulation invariant*
DD correctly simulated by HHH cannot possibly reach its
simulated final state.

Your own answer to A or B:

On 10/9/2026 6:53 AM, Python wrote:

You have established that correct application of the C operational
semantics within HHH produces the recursively simulated sequence:

ÿÿÿ HHH simulates DD
ÿÿÿÿÿÿÿ -> DD calls HHH(DD)
ÿÿÿÿÿÿÿÿÿÿÿ -> HHH simulates DD
ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ -> ...

Agreed.

Peter, I understand perfectly well what I agreed to.

I agreed that IF HHH keeps recursively simulating the HHH call
encountered at DD line 03, then we obtain:

ÿÿ HHH -> DD -> HHH -> DD -> ...

and that recursive simulation does not naturally return.

The following is all within the context of DD and
HHH being correctly simulated by HHH while this
correct simulation continues:

<begin>
Your own "does not naturally return" means HHH
does not reach its own "return" statement.

When HHH does not reach its own "return" statement
then DD never receives a value from HHH.

When DD never receives a value from HHH then
DD never reaches its own "return" statement.

When DD never reaches its own "return" statement
then the correctly simulated DD never stops running.
<end>

I have agreed to that approximately seventeen fucking times now.

What I did NOT agree to is your next inference:

ÿÿ therefore the finite source-code DD specifies
ÿÿ a computation that does not halt.

Forgetting (or never knowing) that halting is
defined as reaching a "return" statement final state.

Those are not the same proposition.

And your own words expose the exact problem:

keeps repeating unless aborted

UNLESS ABORTED.

There it is.

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider HHH correctly simulates its
input DD until HHH correctly determines that its simulated DD
*would never stop running unless aborted* then

Thus never halting while being correctly simulated.
AKA simulated DD would never stop running unless aborted
while being correctly simulated.

That is the entire fucking issue.

You claim that HHH detects this recursion, ABORTS the simulation, and returns 0.

Fine.

No you are not paying close enough attention. You did
not bother to notice that this condition has been met:

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider HHH correctly simulates its
input DD until HHH correctly determines that its simulated DD
*would never stop running unless aborted* then

Then tell me what happens to the HHH call at line 03 of the DD being simulated.

You did not pay close enough attention. that once
the condition has been met that DD is legitimately killed.

HHH can abort its simulation of DD and correctly report that
DD specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

There are only two possibilities.

FIRST:

The simulated call

ÿÿ HHH(DD)

at DD line 03 returns the 0 that you claim HHH returns.

Then the C operational semantics are embarrassingly simple:

ÿÿ 03ÿÿ int Halt_Status = HHH(DD);

gives:

ÿÿ Halt_Status = 0

then:

ÿÿ 04ÿÿ if (Halt_Status)

is false, so line 05 is skipped, and:

ÿÿ 06ÿÿ return Halt_Status;

is reached.

DD HALTS.

SECOND:

The simulated call to HHH at line 03 does NOT return 0 because you
continue recursively simulating:

ÿÿ HHH -> DD -> HHH -> DD -> ...

Fine.

Then your simulation never produces the alleged 0 for that call.

In that case you have not shown that HHH returns 0 there.

You have shown that YOUR METHOD FOR COMPUTING HHH'S ANSWER DOES NOT TERMINATE.

That is the dilemma you have been running away from for weeks.

And no, my October 9 statement does not answer it for you.

Read it:

correct application of the C operational
semantics within HHH produces the recursively simulated sequence

Yes.

THE RECURSIVELY SIMULATED SEQUENCE.

I agreed about the behavior of your recursive simulation.

I did NOT say:

ÿÿ therefore DD does not halt.

In fact, a few days ago when I explicitly pointed this out, you answered:

Of course it does not say this.

Remember?

So please stop pretending that my October 9 statement contains words
that YOU HAVE ALREADY ADMITTED IT DOES NOT CONTAIN.

Your latest post actually makes the defect beautifully explicit:

ÿÿ "keeps repeating unless aborted"

If it is NOT aborted:

ÿÿ HHH does not finish deciding DD.

If it IS aborted and 0 is returned to the simulated DD:

ÿÿ DD reaches line 06 and halts.

Pick your poison.

You do not get a magical third option in which:

ÿÿ HHH aborts and returns 0

but simultaneously:

ÿÿ the DD waiting for that return at line 03
ÿÿ never receives the 0

while you still call the whole thing a correct simulation according to
the C operational semantics.

That is not C semantics.

That is sleight of hand.

You are using the non-return of HHH to prove that DD cannot get past
line 03, while simultaneously using the return of HHH to claim that HHH
is a terminating decider.

You need HHH NOT TO RETURN to prove your conclusion about DD.

You need HHH TO RETURN to call HHH a decider.

You cannot have both.

And THAT -- not some "mental block" of mine -- is the contradiction you
keep hiding behind the word "aborted."

--
Copyright 2026 Olcott

Axiom 1 --- Paraphrase of Emil Post definition of computation
All computations are equivalent to the application of finite
string transformation rules to finite string inputs to derive
any and all outputs.

Axiom 2 --- Correcting the misnomer term-of-the-art: decider
Deciders must apply finite string transformation inference
steps to their finite string inputs that are relevant to the
decision criteria and make their decision on this basis.

*Succinct refutation of the Halting Theorem* https://github.com/plolcott/x86utm/blob/master/README.md


The essence of computation is applying finite
string transformations to finite string inputs
to derive outputs. // paraphrase of Emil Post

My 28 year goal has been to make
"true on the basis of meaning expressed in language"
reliably computable for the entire body of general knowledge.
The complete structure of this system is now defined.

The entire body of knowledge expressed in language is
comprised of two types