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On 6/19/2024 12:43 PM, joes wrote:Am Wed, 19 Jun 2024 08:31:24 -0500 schrieb olcott:On 6/19/2024 4:08 AM, joes wrote:Am Tue, 18 Jun 2024 21:51:56 -0500 schrieb olcott:On 6/18/2024 9:41 PM, Richard Damon wrote:On 6/18/24 10:30 PM, olcott wrote:On 6/18/2024 9:16 PM, Richard Damon wrote:On 6/18/24 1:25 PM, olcott wrote:On 6/18/2024 12:06 PM, joes wrote:
Terminating is a property of the actual machine, and not a
simulation of it.
Your nonsimulator halts even when given nonterminating input.YOUR partial decider makes everything halt, even that which doesn't.
So yes, it can't simulate infinite loops.
When I write an infinite loop, I want it to be interpreted as anYou could say the SIMULATION didn't terminate normally, but youSure you can otherwise interpreters of source-code would be a bogus
can't say the machine didn't or even the Turing Machine
Description, as you could give that exact same TMD to a real UTM
and find out the actual behaviof or the input.
concept.
infinite loop. Your H0 is bogus.
If the simulation of the x86 machine language of the function doesYour H0 is not an interpreter. It aborts infinite loops.
not prove the actual behavior that this finite string specifies then
source-code interpreters are a bogus concept.
I was talking about DDD. It calls H0, which shall halt. Then DDD[The verification that it ... ?]Which doesn't mean the program DDD needs to be abort to have itThe verified that that it does need to be aborted contradicts your
halt.
nonsense to the contrary.
If H0 halts, so does DDD (which only calls it).
returns, having terminated.
Exactly which step of DDD emulated by H0 was emulated incorrectly suchIt was aborted before it reached there.
that this emulation would be complete? AKA DDD emulated by H0 reaches
machine address [000020b3]
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