Re: Anyone that disagrees with this is not telling the truth ---- V4

On 8/19/24 8:06 PM, olcott wrote:

On 8/19/2024 6:08 PM, Richard Damon wrote:

On 8/19/24 8:14 AM, olcott wrote:

On 8/19/2024 5:17 AM, Fred. Zwarts wrote:
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*Everything that is not expressly stated below is*
*specified as unspecified*
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void DDD()
{
   HHH(DDD);
   return;
}
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_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]
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*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)
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X = DDD emulated by HHH∞ according to the semantics of the x86 language
Y = HHH∞ never aborts its emulation of DDD
Z = DDD never stops running
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My claim boils down to this: (X ∧ Y) ↔ Z
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void EEE()
{
   HERE: goto HERE;
}
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HHHn correctly predicts the behavior of DDD the same
way that HHHn correctly predicts the behavior of EEE.
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Which proves that the simulation failed to reach the end. This makes the simulation incomplete and therefore incorrect.
The simulating HHH is programmed to abort and halt. The simulated HHH should behave exactly in the same way, so no cheating with the Root variable is allowed.
The the simulating HHH aborts when the simulated HHH has only one cycle to go, after which it would also abort and halt, but the simulating HHH failed to reach this end.

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I made my claim more precise.
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Remember, you said: Everything that is not expressly stated below is*
specified as unspecified
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Therefore HHHn can NOT correctly emulate DDD past the call HHH instruction, because it doesn't HAVE the instruciton of the PROGRAM DDD (which is what you emulate) since it doesn't have the instruction at 000015D2.
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That they are in the same memory space is entailed
in the same way that the x86 code is not being run
on a rubber ducky is entailed.
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 But not EXPLICITLY stated, so that is a lie.
 

And what is WRONG with running the code on a rubber ducky, it might be powered by Pentium.
 And, if they ARE in the same memory space, then it is DDDn not DDD, as there are each different by the memory that came with them.
 Sorry, you are just caught out in your lie and stupdity.
 You just don't knunderstand what you are talking about.
 

The contents of the memory at 000015D2 can not be accessable to HHHn, as the input is described as DDD and not DDDn, so the input doesn't change between instances, and thus CAN'T contain that memory that changes, and thus is not valid to be part of the input.
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Thus we also have that HHH∞ can not exist, so both your premises just fail to be possible.
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Sorry, you are just repeating your error because apparently you just can't learn.
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