On 2021-11-04 15:20, olcott wrote:
On 11/4/2021 3:09 PM, Malcolm McLean wrote:
On Thursday, 4 November 2021 at 17:41:31 UTC, André G. Isaak wrote:
You need to understand the PO is working with x86 code, not Turing
(1) The above 14 simulated lines are a correct pure simulation of the >>>>> input to H(P,P) for every possible encoding of simulating haltWhat exactly does 'every possible encoding of simulating halt
decider H.
decider H'
even mean?
machines.
So P does not contain an embedded near copy of H, as in Linz's scheme,
but a call to H.
Now at first sight, that doesn't matter. But it allows for a mental
separation
between "the input" (the simple little driver which calls H and
inverts behaviour
when it gets the result) and H itself (the halt decider).
If we replace the call to H by a call to another simulating halt
decider, have
we changed anything now?
I think that you have this correctly.
My current proof applies to every possible encoding of H thus
eliminating any need to see the encoding of any specific H.
Again, I ask, what do you mean by an 'encoding of H'?
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