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On 7/20/24 5:21 PM, olcott wrote:You know that you are lying about this as you admit below:On 7/20/2024 4:06 PM, joes wrote:Wrong.Am Sat, 20 Jul 2024 15:05:53 -0500 schrieb olcott:>On 7/20/2024 2:50 PM, Richard Damon wrote:>On 7/20/24 3:09 PM, olcott wrote:On 7/20/2024 2:00 PM, Fred. Zwarts wrote:Op 20.jul.2024 om 17:28 schreef olcott:You missed a couple details:(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.
A terminating input shouldn't be aborted, or at least not classified
as not terminating. Terminating inputs needn't be aborted; they and the
simulator halt on their own.
>Pretty much.And when it aborts, the simulation is incorrect. When HHH aborts andSo you are trying to get away with saying that no HHH ever needs to
halts, it is not needed to abort its simulation, because it will halt
of its own.
abort the simulation of its input and HHH will stop running?I thought they all halt after a finite number of steps?It is the fact that HHH DOES abort its simulation that makes it notNo stupid it is not a fact that every HHH that can possibly exist aborts
need to.
its simulation.
>
void DDD()
{
HHH(DDD);
return;
}
>
DDD correctly simulated by pure function HHH cannot
possibly reach its own return instruction.
>
It may be that the simulation by HHH never reaches that point,
but if HHH aborts its simuliaton and returns (as required for it to be a decider) then the behavior of DDDSimulated by HHH is to Die, stop running, no longer function.
call HHH(DDD), for that HHH toWhy condemn yourself to Hell over this?
partially emulate DDD, and (after the point in that DDD that it emulated) abort that emulation and return and thus DDD will return.--
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