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On 4/3/2025 5:46 PM, Richard Damon wrote:WHich is irrelevent, as HHH to be a decider must abort and return some answer, and thus DDD is halting.On 4/3/25 6:37 PM, olcott wrote:void DDD()On 4/2/2025 9:35 PM, Richard Damon wrote:>On 4/2/25 10:15 PM, olcott wrote:>On 4/2/2025 8:52 AM, joes wrote:>Am Tue, 01 Apr 2025 21:37:40 -0500 schrieb olcott:>On 4/1/2025 8:13 PM, Richard Damon wrote:>On 4/1/25 7:35 PM, olcott wrote:On 4/1/2025 5:36 AM, Richard Damon wrote:On 3/31/25 10:19 PM, olcott wrote:*finiteYou already admitted that you are lying about this.Right, the DDD who's simulation is stopped hasn't shown non-haltingBut DDD doesn't prevent its own terminatation, as it calls an HHHYou know that DDD stopping running and DDD reaching its final halt
that WILL abort its emulation and return and answer.
>
state are not the same thing you damned liar.
>
behavior, just not-yet-halted.
>
DDD emulated by HHH for an infinite number of steps never reaches its
final halt state.
>HHH sees this in one recursive emulation of DDD.It must also return the right value.
>
*Simulating termination analyzer Principle*
It is always correct for any simulating termination analyzer to stop
simulating and reject any input that would otherwise prevent its own
termination. The only rebuttal to this is rejecting the notion that
deciders must always halt.
>
By process of elimination and by the above criteria
we can determine that not stopping the emulation
causes HHH to never halt. This only leaves stopping
the emulation that semantically entails that the
input must be rejected.
>
No, because HHH must do what HHH does.
>
So I am tong because you claim that HHH
does not do what it does?
>
The problem is that your logic claims that HHH doesn't do what it does, which is just emulate its inputpartially and them return.
>
{
HHH(DDD);
return;
}
Do you really think that anyone knowing the C
programming language is too stupid to see that
DDD simulated by HHH cannot possibly return?
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