XPost: comp.theory, sci.logic, sci.math

On 1/31/2022 6:41 PM, Richard Damon wrote:

On 1/31/22 3:24 PM, olcott wrote:

On 1/31/2022 2:10 PM, Ben wrote:

On 1/31/2022 8:06 AM, olcott wrote:

On 1/30/2022 8:20 PM, Richard Damon wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ >>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to H^.Qy/H^.Qn >>>>>>> iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H in
the same way that (5,3) is syntactically specified as an input to
Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to embedded_H in >>>>>> the same way that (1,2) is NOT syntactically specified as an input >>>>>> to Sum(5,3)

Right, but perhaps you don't understand that from you above
statement the right answer is based on if UTM(<H^>,<H^>) Halts
which by the definition of a UTM means if H^ applied to <H^> Halts.

The biggest reason for your huge mistakes is that you cannot stay
sharply focused on a single point. It is as if you either have
attention deficit disorder ADD or are addicted to methamphetamine.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ >>>>  >>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and >>>> Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on to the
points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H embedded
at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must give is
based on the behavior of H^ applied to <H^> BECAUSE OF THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Any moron knows that a function is only accountable for its actual inputs.

embedded_H is only accountable for the actual behavior specified by its
actual inputs ⟨Ĥ⟩ ⟨Ĥ⟩.

Since the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ at Ĥ.qx cannot possibly transition to
its final state: ⟨Ĥ⟩.qn we know that this input never halts even if it stops running.

computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234)



--
Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

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