olcott <NoOne@NoWhere.com> writes:
On 12/25/2021 7:27 PM, Ben Bacarisse wrote:<cut diversions>
olcott <NoOne@NoWhere.com> writes:
On 12/25/2021 2:07 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 12/24/2021 8:50 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 12/24/2021 3:29 PM, Ben Bacarisse wrote:And, much to my surprise, you are clear that you do indeed claim the >>>>>>> second of these to be true!
olcott <NoOne@NoWhere.com> writes:
On 12/23/2021 7:18 PM, Ben Bacarisse wrote:
I will stick to symbols. So you think that one or both of >>>>>>>>>>>
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy but H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊬* H.qy
or
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn but H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊬* H.qn
is possible?
<cut>If so, you are wrong. If you don't, you agree with this: >>>>>>>>>>>
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy if, and only if, H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn if, and only if, H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy
The copy of H at Ĥ.qx... which is applied to ⟨Ĥ⟩ ⟨Ĥ⟩ ...
correctly decides that its input never halts.Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn if, and only if, H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts. >>>>>>> I.e.
but
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
which is obviously absurd. It's logically absurd, but even if you >>>>>>> prevent logical deduction by removing the annotations (as you repeatedly
do), it's still patently absurd to claim that the same state transition >>>>>>> function and input results in different state transitions.
At least you are not trying to hide anything. You explicitly claim an >>>>>>> absurdity. No one can take this seriously.
H.q0 wM w ⊢* H.qy // iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn // iff UTM(wM, w) does not halt
You are frantically throwing up diversions. You claimed, above, that two identical sets of states and transitions can give rise to different configuration sequences:
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
and, a little less vaguely:
"The copy of H at Ĥ.qx correctly decides that its input never halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts."
Formally, in the context of Linz's H/Ĥ you assert that:
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn but H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
You don't want to address this mistake, but it's the elephant in the
room. If TMs don't act solely on the basis of their tape and state transition functions, then they are arbitrary and useless.
It would be dishonest to keep posting other stuff in an attempt to try t
get people to forget you ever said this. You must address it. But then
I think you'd have to accept the proof.
olcott <NoOne@NoWhere.com> writes:
On 12/26/2021 7:44 PM, Ben Bacarisse wrote:
The copy of H and H must give rise to exactly the same transitions.
Simply stating otherwise does not address this basic misunderstanding.
Maybe another copy of H can make the tea?
I think that your confusion may be that: H simulates Ḧ applied to ⟨Ḧ⟩
calls embedded_H simulates ⟨Ḧ⟩ ⟨Ḧ⟩ aborts its simulation transitions
to Ḧ.qn causing H to transition to H.qy IS A SINGLE COMPUTATION.
I am not confused. Two identical TMs[1] will always go through the same state transitions when give identical inputs.
[1] I should really qualify this by referring to isomorphic TMs, but
that would probably confuse you even more. But I dare say that soon you
will pull the "they are not identical" line because you do that sort of
thing when you are stuck.
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