Re: Anyone that claims this is not telling the truth

On 8/17/24 3:00 PM, olcott wrote:

On 8/17/2024 1:50 PM, Richard Damon wrote:

On 8/17/24 2:39 PM, olcott wrote:

 On 8/17/2024 7:29 AM, olcott wrote:
 > void DDD()
 > {
 >    HHH(DDD);
 > }
 >
 > _DDD()
 > [00002172] 55         push ebp      ; housekeeping
 > [00002173] 8bec       mov ebp,esp   ; housekeeping
 > [00002175] 6872210000 push 00002172 ; push DDD
 > [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
 > [0000217f] 83c404     add esp,+04
 > [00002182] 5d         pop ebp
 > [00002183] c3         ret
 > Size in bytes:(0018) [00002183]
 >
 > *It is a basic fact that DDD emulated by HHH according to*
 > *the semantics of the x86 language cannot possibly stop*
 > *running unless aborted* (out of memory error excluded)
 

>
Thus DDD has direct access to HHH in this shared memory space.
>

>
And thus ALL of memory is part of the input,

 Any additional details have no effect what-so-ever on my claim.
 

Suure it does.
Since your argument tries to say that since DDD is the same to all of them, so its the behavior.
You are just admitting to being a LIAR.
As I point out, all your HHH's that abort their emulation of DDD will create a DDD that reaches its final state, just not within the emulation done by HHH.
If you restrict yourself to just HHHs that DO a complete and correct emulation of the input, you don't get any that are the deciders you need them to be.