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On 7/21/2024 9:54 AM, joes wrote:But since DDD needs to be given as a full progran, it means that every input DDD is DIFFERENT so all you have shown is that the aborted simulations of the DDD that callAm Sun, 21 Jul 2024 09:34:57 -0500 schrieb olcott:None-the-less we can examine the exhaustively complete setOn 7/21/2024 9:24 AM, joes wrote:>Am Sun, 21 Jul 2024 08:08:53 -0500 schrieb olcott:On 7/21/2024 6:37 AM, Richard Damon wrote:On 7/21/24 12:15 AM, olcott wrote:A deterministic program can't change. It was always going to be aborted.The behavior of emulated DDD after it has been aborted changes the(b) We know that a decider is not allowed to report on the behaviorThat IS exactly the input.
computation that itself is contained within. Deciders only take finite
string inputs. They do not take executing processes as inputs. Thus
HHH is not allowed to report on the behavior of this int main() {
DDD(); }.
behavior of the directly existed DDD.
>
of every HHH/DDD pair that can possibly exist and find that
all of the HHH instances that never abort their simulation
of DDD never stop running.
And thus all the copies of this HHH will return and thus ALL copies of that DDD will halt.When the second call of what would otherwise be infinite recursion isThe second call stops simulating just like all others.
required to be aborted to prevent the infinite execution of the first
call this proves that HHH(DDD)==0 is correct even though the directly
executed DDD() halts.
>HHH is not an UTM.Unless you think the idea of UTMs is wrong-headed nonsense the behaviorTherefore we map the finite string input to HHH(DDD) to the behaviorThe basis is the direct behaviour.
that it species on the basis of DDD correctly emulated by any pure
function HHH that can possibly exist.
of DDD correctly emulated by HHH determines the actual behavior
specified by the input to HHH(DDD).
>
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