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

On 5/11/25 12:26 PM, olcott wrote:

On 5/11/2025 4:25 AM, Mikko wrote:

On 2025-05-10 15:25:16 +0000, olcott said:
>

On 5/10/2025 2:33 AM, Mikko wrote:

On 2025-05-09 16:25:12 +0000, olcott said:
>

void DDD()
{
   HHH(DDD);
   return;
}
>
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)

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

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

>
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.
>

 _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]
 There are seven instructions in DDD.
When DDD calls HHH(DDD) then HHH emulates
itself emulating DDD.
 

Then either HHH is just wrong, violating the definition of the input, or isn't a pure function, and thus can use my static hack to correct emulate the input to the final state.
Sorry, you are just showing that you are lying by equivocation.