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On 7/22/2024 2:57 AM, Mikko wrote:The definition of a Turing machine does not say that a Turing machineOn 2024-07-21 13:34:40 +0000, olcott said:Turing machines never take actual Turing machines as inputs.
On 7/21/2024 4:34 AM, Mikko wrote:No, we don't. There is no such prohibition.On 2024-07-20 13:11:03 +0000, olcott said:(b) We know that a decider is not allowed to report on the behavior
On 7/20/2024 3:21 AM, Mikko wrote:You are the lying one.On 2024-07-19 14:08:24 +0000, olcott said:*Because this is true I don't understand how you are not simply lying*
When we use your incorrect reasoning we would concludeYou and your HHH can reason or at least conclude correctly about
that Infinite_Loop() is not an infinite loop because it
only repeats until aborted and is aborted.
Infinite_Loop but not about DDD. Possibly because it prefers to
say "no", which is correct about Infinte_loop but not about DDD.
int main
{
DDD();
}
Calls HHH(DDD) that must abort the emulation of its input
or {HHH, emulated DDD and executed DDD} never stop running.
If HHH(DDD) abrots its simulation and returns true it is correct as a
halt decider for DDD really halts.
computation that itself is contained within.
They only take finite strings as inputs and an actual executing
Turing machine is not itself a finite string.
Therefore It is not allowed to report on its own behavior.Anyway, that does not follow. The theory of Turing machines does not
Another different TM can take the TM description of thisIf a Turing machine can take a description of a TM as its input
machine and thus accurately report on its actual behavior.
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