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Am Sun, 02 Jun 2024 10:02:54 -0500 schrieb olcott:Introduction to the Theory of Computation, by Michael SipserOn 6/2/2024 4:36 AM, joes wrote:The simulation is incorrect if it stops at that point and the simulatedAm Sat, 01 Jun 2024 17:37:28 -0500 schrieb olcott:>On 6/1/2024 5:30 PM, Richard Damon wrote:>On 6/1/24 5:27 PM, olcott wrote:On 6/1/2024 4:15 PM, Richard Damon wrote:On 6/1/24 4:35 PM, olcott wrote:On 6/1/2024 3:29 PM, Richard Damon wrote:On 6/1/24 12:46 PM, olcott wrote:On 6/1/2024 11:33 AM, Richard Damon wrote:On 6/1/24 12:18 PM, olcott wrote:On 6/1/2024 11:08 AM, Richard Damon wrote:On 6/1/24 11:58 AM, olcott wrote:On 6/1/2024 10:46 AM, Richard Damon wrote:On 6/1/24 10:00 AM, olcott wrote:Not simulating an infinite number of steps of infinite recursion is>Every DD correctly simulated by any HH of the infinite set of HH/DD>
pairs that match the above template never reaches past its own
simulated line 03 in 1 to ∞ steps of correct simulation of DD by HH.
But since the simulation was aborted,
*The above never mentions anything about any simulation being aborted*
incorrect. You always forget this requirement: the simulation must be
complete.
When HH correctly simulates N steps of DD it is incorrect to say that
these N steps were incorrectly simulated.
machine is not in a final state, even if it was correct up to that point.
It also matters what steps where simulated, not only if each was correct.
There are an infinite number of different HH/DD pairs specified by thatI see only a single DD. All H that stop simulating D before it reaches aWhat are the sets of HH and DD? I thought they were concrete machines.The infinite set of every HH/DD pair where HH correctly simulates 1 or
>
more steps of DD is the infinite set that I am referring to.
final state are already wrong.
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