On 1/3/22 10:24 AM, olcott wrote:
On 1/3/2022 9:18 AM, Richard Damon wrote:
LIAR !!!
On 1/3/22 9:25 AM, olcott wrote:
Revised Linz H halt deciding criteria (My criteria Ben's notation)
H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
The above means that the simulating halt decider H bases its halt
status decision on the behavior of the pure UTM simulation of its
input.
H examines this behavior looking for infinite behavior patterns.
When H detects an infinite behavior pattern it aborts the simulation
of its input and transitions to H.qn.
This pattern does not exist as a finite pattern.
Proved, and accepted by failure to rebut.
Mesage ID <FOnzJ.162569$np6.119786@fx46.iad>
Date: 2021-12-30 19:31:49 GMT
Subject: Re: Concise refutation of halting problem proofs V42
[compute the mapping]
FAIL.
Infinite behavior patterns are cases where the the pure UTM
simulation of the input would never reach the final state of this
input.
For simplicity we will refer to the copy of Linz H at Ĥ.qx embedded_H. >>>>
Simplified syntax adapted from bottom of page 319:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
When embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩ these steps would keep repeating:
Ĥ copies its input ⟨Ĥ⟩ to ⟨Ĥ⟩ then embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩...
This shows that the input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ would never reach
its final state thus conclusively proving that this input never halts. >>>> This enables embedded_H to correctly transition to Ĥ.qn.
WRONG.
PROVE IT, or YOUR the LIAR.
I have shown my proof, which you have failed to give a rebuttal that
actually tries to rebut it.
IF embedded_H doesn't abort, then H never gets to Qn as claimed
If embedded_H does abort and go to Qn, then H^ also goes to Qn and Halts.
H and embedded_H must be the same algorithm with the same input so must behave the same.
WHAT IS THE ERROR?, speak now with REAL proof or admit you are the LIAR.
On 1/3/22 11:34 AM, olcott wrote:
On 1/3/2022 9:28 AM, Richard Damon wrote:
On 1/3/22 10:24 AM, olcott wrote:
On 1/3/2022 9:18 AM, Richard Damon wrote:
LIAR !!!
On 1/3/22 9:25 AM, olcott wrote:
Revised Linz H halt deciding criteria (My criteria Ben's notation) >>>>>> H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
The above means that the simulating halt decider H bases its halt
status decision on the behavior of the pure UTM simulation of its
input.
H examines this behavior looking for infinite behavior patterns.
When H detects an infinite behavior pattern it aborts the
simulation of its input and transitions to H.qn.
This pattern does not exist as a finite pattern.
Proved, and accepted by failure to rebut.
Mesage ID <FOnzJ.162569$np6.119786@fx46.iad>
Date: 2021-12-30 19:31:49 GMT
Subject: Re: Concise refutation of halting problem proofs V42
[compute the mapping]
FAIL.
Infinite behavior patterns are cases where the the pure UTM
simulation of the input would never reach the final state of this
input.
For simplicity we will refer to the copy of Linz H at Ĥ.qx
embedded_H.
Simplified syntax adapted from bottom of page 319:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
When embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩ these steps would keep repeating:
Ĥ copies its input ⟨Ĥ⟩ to ⟨Ĥ⟩ then embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩...
This shows that the input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ would never reach
its final state thus conclusively proving that this input never
halts.
This enables embedded_H to correctly transition to Ĥ.qn.
WRONG.
PROVE IT, or YOUR the LIAR.
I have shown my proof, which you have failed to give a rebuttal that
actually tries to rebut it.
IF embedded_H doesn't abort, then H never gets to Qn as claimed
If embedded_H does abort and go to Qn, then H^ also goes to Qn and
Halts.
embedded_H is only accountable for mapping the behavior of the pure
simulation of its input ⟨Ĥ⟩ ⟨Ĥ⟩ to an accept / reject state.
Right, and to correctly answer the input <H^> <H^> then H/embedded_H
must go to the state that matches the behavior of the Computation of H^ applied to <H^>.
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