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

On 2024-07-14 14:50:47 +0000, olcott said:

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.

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.
--
Mikko