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

On 7/14/2024 5:09 AM, Mikko wrote:

On 2024-07-12 14:56:05 +0000, olcott said:
 

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]
>
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
>
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.
>
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.

 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.
 

DDD is the exact same fixed constant finite string that
always calls HHH at the same fixed constant machine
address.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
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