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On 11/3/2024 3:59 PM, joes wrote:A vaceous statement, since your HHH never does a complete emulation that can show "never halts"Am Sun, 03 Nov 2024 12:33:44 -0600 schrieb olcott:On 11/3/2024 12:20 PM, Richard Damon wrote:>On 11/3/24 9:39 AM, olcott wrote:You understood it wrong previously.Yes this is exactly correct. I don't understand why you keep disagreeingThe finite string input to HHH specifies that HHH MUST EMULATE ITSELFRight, and it must CORRECTLY determine what an unbounded emulation of
emulating DDD.
that input would do, even if its own programming only lets it emulate a
part of that.
with your own self this.
>DDD emulated by HHH specifies that HHH will emulate itself emulatingThe finite string input to HHH1 specifies that HHH1 MUST NOT EMULATEBut the semantics of the string haven't changed, as the string needs to
ITSELF emulating DDD.
contain all the details of how the machine it is looking at will work.
DDD.
DDD emulated by HHH1 specifies that HHH1 will NOT emulate itself
emulating DDD.And here we have you cardinal mistake: this case requires DDD to callDDD correctly emulated by HHH never halts and the
its own emulator. We are interested in that program which is constructed
from it; it doesn't exist on its own but depends on HHH/HHH1.
Usually a program is specified by its code, including everything that
it calls. But even HHH1 cannot simulate EEE(){HHH1(EEE);}.
>
exact same thing goes for Linz ⟨Ĥ⟩ simulated by Linz
embedded_H:
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
It seems a little nutty that you refer to a non-existent EEE.
Unless HHH rejects its input DDD as non halting the executed DDD neverYou have cause and effect backwards.
stops running. This itself proves that HHH is correct and that DDD is
not the same instance as the one that HHH rejected.
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