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On 7/20/2024 2:50 PM, Richard Damon wrote:But every HHH that is a decider does.On 7/20/24 3:09 PM, olcott wrote:No stupid it is not a fact that every HHH that can possiblyOn 7/20/2024 2:00 PM, Fred. Zwarts wrote:>Op 20.jul.2024 om 17:28 schreef olcott:>void DDD()>
{
HHH(DDD);
}
>
int main()
{
DDD();
}
>
(a) Termination Analyzers / Partial Halt Deciders must halt
this is a design requirement.
>
(b) Every simulating termination analyzer HHH either
aborts the simulation of its input or not.
>
(c) Within the hypothetical case where HHH does not abort
the simulation of its input {HHH, emulated DDD and executed DDD}
never stop running.
>
This violates the design requirement of (a) therefore HHH must
abort the simulation of its input.
And when it aborts, the simulation is incorrect. When HHH aborts and halts, it is not needed to abort its simulation, because it will halt of its own.
So you are trying to get away with saying that no HHH
ever needs to abort the simulation of its input and HHH
will stop running?
>
It is the fact that HHH DOES abort its simulation that makes it not need to.
exist aborts its simulation.
void DDD()No, you show that the EMULATION BY HHH never reaches that point, not that DDD doesn't reach that point.
{
HHH(DDD);
return;
}
It *is* a fact that no DDD correctly simulated by any
pure function HHH ever reaches its own return instruction.
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