Re: ZFC solution to incorrect questions: reject them

On 3/15/2024 2:14 PM, Richard Damon wrote:

On 3/15/24 12:00 PM, olcott wrote:

On 3/15/2024 1:38 PM, Richard Damon wrote:

On 3/15/24 7:41 AM, olcott wrote:

On 3/15/2024 5:44 AM, Mikko wrote:

On 2024-03-15 01:12:19 +0000, olcott said:
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On 3/14/2024 8:06 PM, Richard Damon wrote:

On 3/14/24 4:45 PM, olcott wrote:

On 3/14/2024 5:37 PM, Richard Damon wrote:

On 3/14/24 3:04 PM, olcott wrote:

On 3/14/2024 4:55 PM, Richard Damon wrote:

On 3/14/24 1:59 PM, olcott wrote:

On 3/14/2024 3:54 PM, Richard Damon wrote:

On 3/14/24 1:26 PM, olcott wrote:

On 3/14/2024 3:20 PM, Richard Damon wrote:

On 3/14/24 12:32 PM, olcott wrote:

On 3/14/2024 12:33 PM, Richard Damon wrote:

On 3/13/24 4:04 PM, olcott wrote:

On 3/13/2024 5:43 PM, Richard Damon wrote:

On 3/13/24 2:54 PM, olcott wrote:

On 3/13/2024 4:39 PM, Richard Damon wrote:

On 3/13/24 1:52 PM, olcott wrote:

On 3/13/2024 12:52 PM, Richard Damon wrote:

On 3/13/24 10:08 AM, olcott wrote:

On 3/13/2024 11:44 AM, immibis wrote:

On 13/03/24 04:55, olcott wrote:

On 3/12/2024 10:49 PM, Richard Damon wrote:

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Not quite. It always gets the wrong answer, but only one of them for each quesiton.
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They all gets the wrong answer on a whole class of questions

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Wrong. You said. yourself. that H1 gets the right answer for D.
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Since it is a logical impossibility to determine the truth
value of a self-contradictory expression the requirement
for H to do this is bogus.

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Shows you are just a LIAR, as there IS a truth value to the expression that is the requirment for ANY SPECIFIC H.
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*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
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There is no mapping from H(D,D) to Halts(D,D) that exists.
This proves that H(D,D) is being asked an incorrect question.
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Why, because it is NOT a LIE.
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You don't even know the definiton of an incorrect question.
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I invented it so I get to stipulate its meaning.
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ

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Nope, common technical term.
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Cite a source.
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The fact that there DOES exist a mapping Halt(M,d) that maps all Turing Machines and there input to a result of Halting / Non-Halting for EVERY member of that input set, means tha Halts is a valid mapping to ask a decider to try to decider.
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That part is true.
Likewise when you ask a man that has never been married:
Have you stopped beating tour wife?
There are some men that have stopped beating their wife.

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Right, because that question include a presumption of something not actually present.
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Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO
thus the question is incorrect for all unmarried men.
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Although there is a mapping from some TM/input pairs to YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
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Except that the mapping requested is about the INPUTS to H, not H itsef.
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In order to see that it is an incorrect question we must examine
the question in detail. Making sure to always ignore this key detail
<is> cheating.
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Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))

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Which isn;t the question at all, so you are just shown to be a stupid liar.
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The QUESTION is:
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Does the machine and input described by this input, Halt when run?

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The question posed to Ĥ.H has no correct answer, thus not the
same question at all.

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But it DOES.

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Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
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I didn't say there was.
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Then you understand that each question posed to each Ĥ.H in the
above set has no correct answer only because each of these answers
are contradicted by the machine that H is contained within.
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No, YOU don't understand that the IS a correct answer, just not the one that H (or H^.H ) happens to give.
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Then show me which contradicted answer is correct.
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If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn was the right answer.

*The strawman deception is all that you have*
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Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
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*The answer must come from elements of the above set*

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Is a false claim about a strawman deception really the best you can say?
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The above are the program/input pairs such that every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
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So?
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That doesn't mean they are the set that the answer to the ACTUAL QUESTION needs to come from.
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You are just proving your stupidity and duplicity.

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Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
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Credit goes to you for finding the loophole in Carol's original
question: Can anyone correctly answer “no” to this question?
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Carol can correctly answer that question with any word that is
synonymous with "no".
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Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question?
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The fact that anyone besides Carol can correctly answer that
question with a NO and Carol cannot possibly correctly answer
that question proves that it is a different question when posed
to Carol than when posed to anyone else.

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Which is IRRELEVENT to the Halting Question, as it is a purely objective question.
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The behavior of the input is INDEPENDENT of the decider looking at it.
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Note, a given H^ is built on a given H, and no other, but can be given to any decider to answer, and the correct answer will be the same irrespective of you ask. Some will give the right answer, and some will give the wrong answer. The fact that that H is in the latter doesn't make the question subjective.
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The only way to make the Halting Question subjective is to try to redefine it so the input changes with who you ask, but it doesn't.
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The changing H^ to match the H only happens in the Meta, where we prove that we can find an H^ that any H will get wrong, but each of those are SEPERATE Halting question (not all one question) and each of those seperate questions have a correct answer.
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Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
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Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such as H1
(that is not contradicted) can determine a correct answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
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Nope.
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The Question doesn't refer to H at all.
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 The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question?

 Thus changing the meaning of the question
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
*in the exact same way that the meaning of this question*
 Can Carol correctly answer “no” to this [yes/no] question?
is changed when it is posed to Carol.