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

On 7/17/24 9:04 AM, olcott wrote:

On 7/17/2024 1:52 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.
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_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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The semantics does not prove. Only a proof would prove.
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 Nothing besides the semantics of English proves that
a kitten is not any type of 15 story office building.
 _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]
 DDD emulated by HHH according to the semantic meaning of
its x86 instructions never stop running unless aborted.
 

Wrong, EVERY DDD that is calls an HHH that any copy of it returns from the call HHH(DDD) to any caller makes a DDD that halts.
Yes, the HHH can never reach that state in its emulation, but DDD gets there.
You are just proving you are a stupid liar that doesn't know what he is talking about.