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On 3/9/2025 10:48 AM, joes wrote:And thus NOTHING You say has anything to do with the Halting Problem and is just part of your giant fraud based on lying about working in the system becuase you have admitted that you have changed core defintions, and thus everything you have done is just a lying fraud.Am Sun, 09 Mar 2025 09:37:12 -0500 schrieb olcott:If you totally understood what I mean you wouldOn 3/9/2025 9:28 AM, dbush wrote:>On 3/9/2025 10:26 AM, olcott wrote:On 3/9/2025 9:11 AM, dbush wrote:On 3/9/2025 10:08 AM, olcott wrote:On 3/9/2025 8:50 AM, dbush wrote:Oh really? I do understand what you mean. What's next? Or do you haveYou cannot possibly understand anything that I say about that until youIn other words, you're disagreeing with a stipulative definition.>When we assume that HHH emulates N steps of DD thenIt is stipulated that a solution to the halting problem perform the
*DD correctly emulated by HHH cannot possibly reach*
*its own "ret" instruction and terminate normally*
*because DD calls HHH(DD) in recursive emulation*
I am not going to address any other point until this point is fully
understood because the other points cannot possibly be understood
until this one is totally understood.
Whether or not and how it applies to the Halting Theorem cannot
possibly be understood at all until after the above words are 100%
totally and perfectly understood.
>
following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
directly
If you went to play head games you can play by yourself.
As you yourself said:
>
after you first understand this:
[spam]
nothing?
>
not have given me such a hard time about it.
When we assume that HHH emulates N steps of DD then
DD correctly emulated by HHH cannot possibly reach
its own "ret" instruction and terminate normally
because DD calls HHH(DD) in recursive emulation.
If you do not agree that HHH it correct to reject
DD as non-terminating then you do not understand.
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