Re: Computing the mapping from the input to HHH(DD) --- REFUTES INCORRECT REQUIREMENTS

On 5/12/25 11:07 AM, olcott wrote:

On 5/12/2025 3:04 AM, Mikko wrote:

On 2025-05-11 16:26:44 +0000, olcott said:
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On 5/11/2025 4:25 AM, Mikko wrote:

On 2025-05-10 15:25:16 +0000, olcott said:
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On 5/10/2025 2:33 AM, Mikko wrote:

On 2025-05-09 16:25:12 +0000, olcott said:
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void DDD()
{
   HHH(DDD);
   return;
}
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When 1 or more statements of DDD are correctly
simulated by HHH then this correctly simulated
DDD cannot possibly reach its own “return statement”.
(final halt state)

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That one or more statements of DDD are correctly simulated does not
mean that DDD is correctly simulated.

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It is stipulated that when one or more statements
of DDD are correctly simulated that one or more
statements of DDD are correctly simulated.

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Thera are only two statements in DDD. HHH does not correctly emulate
the first one, which is a call to HHH, and not at all the second one,
which is the final return.

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_DDD()
[0000219e] 55             push ebp
[0000219f] 8bec           mov ebp,esp
[000021a1] 689e210000     push 0000219e // push DDD
[000021a6] e843f4ffff     call 000015ee // call HHH
[000021ab] 83c404         add esp,+04
[000021ae] 5d             pop ebp
[000021af] c3             ret
Size in bytes:(0018) [000021af]
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There are seven instructions in DDD.
When DDD calls HHH(DDD) then HHH emulates
itself emulating DDD.

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No, there are seven instructions in a particular translation of DDD.

 Not at all. The above sequence of machine code bytes
is named DDD. Every other string of bytes is off topic.
The x86 machine code is the final form.

And the above string of bytes are NOT a program, and can NOT be correctly emulated by a pure emulator past the call instruction.
Sorry, but you just are showing that you don't care about truth, but just want to repeat your lies.

 

But that is irrelevant. My comment was to a message that did not use
the word "instruction" at all.
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