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On 8/3/2024 11:12 AM, Richard Damon wrote:No, you are just proving you are incapable of learning.On 8/3/24 12:03 PM, olcott wrote:Great! Finally.On 8/3/2024 10:33 AM, Richard Damon wrote:>On 8/3/24 10:26 AM, olcott wrote:>On 8/3/2024 9:04 AM, Fred. Zwarts wrote:>Op 03.aug.2024 om 15:50 schreef olcott:>On 8/3/2024 3:14 AM, Fred. Zwarts wrote:>Op 02.aug.2024 om 22:57 schreef olcott:>Who here is too stupid to know that DDD correctly simulated>
by HHH cannot possibly reach its own return instruction?
>
void DDD()
{
HHH(DDD);
return;
}
>
Which proves that the simulation is incorrect.
When are you going to understand that you are not allowed
to disagree with the semantics of the x86 language?
>
>
I do not disagree.
When are you going to understand that it is a deviation of the semantics of the x86 language to skip instructions of a halting program,
HHH(DDD) simulates DDD that calls HHH(DDD) to repeat the process.
>
If it does this an infinite number of times the simulated DDD
never reaches its own return instruction.
>
If it does this a googolplex number of times the simulated DDD
never reaches its own return instruction.
Nope, the PARTIAL SIMULATION of DDD never reaches the return instruction.
>
For N = 0; while N <= googolplex; N++
N instructions of DDD correctly emulated by HHH[N] never
reach their own "return" instruction final state.
>
∞ instructions of DDD correctly emulated by HHH[∞] never
reach their own "return" instruction final state.
>
Thus any HHH that takes a wild guess that DDD emulated
by itself never halts is always correct.
>
The SIMULATION of DDD never reaches the return instruction.
>
When we understand that the return instruction is halt state
of DDD then DDD correctly simulated by HHH never halts.
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