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Am Sat, 29 Mar 2025 16:46:26 -0500 schrieb olcott:UTM1 simulates D that calls UTM1On 3/29/2025 3:14 PM, dbush wrote:On 3/29/2025 4:01 PM, olcott wrote:A complete simulation of a nonterminating input doesn't halt.It is dishonest to expect non-terminating inputs to complete.We can know that when this adapted UTM simulates a finite number ofAnd therefore does not do a correct UTM simulation that matches the
steps of its input that this finite number of steps were simulated
correctly.
behavior of the direct execution as it is incomplete.
So not an UTM.When UTM1 is a UTM that has been adapted to only simulate a finiteFalse, if the starting function calls UTM and UTM changes, you're2) changing the input is not allowedThe input is unchanged. There never was any indication that the input
was in any way changed.
changing the input.
number of steps
and input D calls UTM1 then the behavior of D simulatedDoesn't matter if it calls it, but if the UTM halts.
by UTM1 never reaches its final halt state.
When D is simulated by ordinary UTM2 that D does not call Then D reaches
its final halt state.
You changed UTM1, which is part of the input D.Changing the input is not allowed.I never changed the input. D always calls UTM1.
thus is the same input to UTM1 as it is to UTM2.
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