XPost: comp.theory, sci.logic, sci.math
On 1/31/2022 10:17 PM, Richard Damon wrote:
On 1/31/22 10:40 PM, olcott wrote:
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)?
Don't know how you get that from what I said.
Any moron knows that a function is only accountable for its actual
inputs.
And the actual input to H is <H^> <H^> which MEANS by the DEFINITION of
the Halting Problem that H is being asked to decide on the Halting
Status of H^ applied to <H^>
No that is not it. That is like saying "by definition" Sum(3,5) is being
asked about Sum(7,8).
Halt decider are deciders thus are only ever accountable for the
properties of their actual inputs.
The function that embedded_H computes is being asked about the actual
behavior of the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ at Ĥ.qx.
--
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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