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On 8/3/2024 12:45 PM, Richard Damon wrote:No, you just don''t know what you are talking about because you lie to yourself.On 8/3/24 12:35 PM, olcott wrote:You said: "doesn't" you did not say: "does"On 8/3/2024 11:32 AM, Richard Damon wrote:>On 8/3/24 12:16 PM, olcott wrote:>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:>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.
>
Great! Finally.
When we understand that the return instruction is halt state
of DDD then DDD correctly simulated by HHH never halts.
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>
The PARTIAL simulation of DDD done by HHH doesn't reach the return instruction.
>
∞ instructions of DDD correctly emulated by HHH[∞] never
reach their own "return" instruction final state.
>
So you are saying that the infinite one does?
>
I said for the HHH's that do a partial simulation it does.
>
You can't even accurately quote your own self?
Every DDD correctly emulated by any HHH for a finite orNope. And you statment is just a incoherent statement, as no partial simulaitoni for a finite number of steps is "correct".
infinite number of steps never reaches its own "return"
halt state.
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