Re: DDD correctly emulated by HHH is correctly rejected as non-halting V2

On 8/1/2024 2:49 AM, Mikko wrote:

On 2024-07-31 17:28:38 +0000, olcott said:
 

On 7/31/2024 2:36 AM, Mikko wrote:

On 2024-07-16 18:18:07 +0000, olcott said:
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On 7/16/2024 2:57 AM, Mikko wrote:

On 2024-07-15 13:43:34 +0000, olcott said:
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On 7/15/2024 3:17 AM, Mikko wrote:

On 2024-07-14 14:50:47 +0000, olcott said:
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On 7/14/2024 5:09 AM, Mikko wrote:

On 2024-07-12 14:56:05 +0000, olcott said:
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We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
>
_DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
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When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
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When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
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The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.

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You should use the indices here, too, e.g., "where 1 to infinity steps of
DDD₁ are correctly emulated by HHH₃" or whatever you mean.
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DDD is the exact same fixed constant finite string that
always calls HHH at the same fixed constant machine
address.

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If the function called by DDD is not part of the input then the input does
not specify a behaviour and the question whether DDD halts is ill-posed.
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We don't care about whether HHH halts. We know that
HHH halts or fails to meet its design spec.
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We are only seeing if DDD correctly emulated by HHH
can can possibly reach its own final state.

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HHH does not see even that. It only sees whther that it does not emulate
DDD to its final state.

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No. HHH is not judging whether or not itself is a correct
emulator. The semantics of the x86 instructions that emulates
prove that its emulation is correct.

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Semantics of x86 language alone doesn't prove anything. Only a detailed
comparison of the emulator code to the x86 semantics may prove that.

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A proof is any sequence of steps such that the conclusion
is a necessary consequence of its basis.

 Only if every "step" is a sentence.
 

Not at all.
My proof is the tautology that proves it is correct.

Your traces are not such sequences.
 

Proving that DDD correctly emulated by HHH matches the
infinite recursion behavior pattern.
(a) The semantics of the x86 language.
(b) the design of HHH provided below.
(c) The definition of infinite recursion provided below.
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*Infinite recursion behavior pattern*
An emulated sequence of instructions that has no conditional
branch instructions in this sequence is exactly repeated when
it calls the same function with the same parameters again.
As long as the called function can be determined to never
return this proves infinite recursion.

 The recursive simulation does not satisfy the "no conditional branch"
requirement so does not match the pattern.
 

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If *simulating halt decider H correctly simulates its input D*
     *until H correctly determines that its simulated D would never*
     *stop running unless aborted* then
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
It freaking meets the above highlighted criteria.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer