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On 8/10/24 3:15 PM, olcott wrote:*Yes this is your ADD*On 8/10/2024 1:58 PM, Richard Damon wrote:But since none of your traces show more that 4, that is a lie, since you haven't been able to establish the HHH itself correctly emulates ANY of the instructions of the program DDD after the call HHH, as everything says it jumps to something other than the correct x86 emulation of the program DDD that it was given.On 8/10/24 2:11 PM, olcott wrote:>On 8/10/2024 12:51 PM, Richard Damon wrote:On 8/10/24 1:37 PM, olcott wrote:>>
void DDD()
{
HHH(DDD);
return;
}
>
*The set of HHH x86 emulators are defined such that*
Each element of this set corresponds to one element of the set
of positive integers indicating the number of x86 instructions
of DDD that it emulates.
*This is the mistake that I corrected*
But, we can overlook that, since you fail otherways.
And it still does. If HHH emulates for a finite number of steps, then returns, then the PROGRAM DDD that calls that HHH will halt.>But every one that emulates for a finite number of steps, and then returns create a halting DDD, so you claim is just disproven.
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