Liste des Groupes | Revenir à s logic |
On 3/15/2024 5:44 AM, Mikko wrote:More specifically, the right answer contradict. Being contracdictOn 2024-03-15 01:12:19 +0000, olcott said:The above are the program/input pairs such that every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
On 3/14/2024 8:06 PM, Richard Damon wrote:Is a false claim about a strawman deception really the best you can say?On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*On 3/14/2024 5:37 PM, Richard Damon wrote:If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn was the right answer.On 3/14/24 3:04 PM, olcott wrote:Then show me which contradicted answer is correct.On 3/14/2024 4:55 PM, Richard Damon wrote:No, YOU don't understand that the IS a correct answer, just not the one that H (or H^.H ) happens to give.On 3/14/24 1:59 PM, olcott wrote:Then you understand that each question posed to each Ĥ.H in theOn 3/14/2024 3:54 PM, Richard Damon wrote:I didn't say there was.On 3/14/24 1:26 PM, olcott wrote:Then tell me which element of:On 3/14/2024 3:20 PM, Richard Damon wrote:But it DOES.On 3/14/24 12:32 PM, olcott wrote:The question posed to Ĥ.H has no correct answer, thus not theOn 3/14/2024 12:33 PM, Richard Damon wrote:Which isn;t the question at all, so you are just shown to be a stupid liar.On 3/13/24 4:04 PM, olcott wrote:In order to see that it is an incorrect question we must examineOn 3/13/2024 5:43 PM, Richard Damon wrote:Except that the mapping requested is about the INPUTS to H, not H itsef.On 3/13/24 2:54 PM, olcott wrote:Cite a source.On 3/13/2024 4:39 PM, Richard Damon wrote:Nope, common technical term.On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning.On 3/13/2024 12:52 PM, Richard Damon wrote:Why, because it is NOT a LIE.On 3/13/24 10:08 AM, olcott wrote:*Lying about me being a liar may possibly cost your soul*On 3/13/2024 11:44 AM, immibis wrote:Shows you are just a LIAR, as there IS a truth value to the expression that is the requirment for ANY SPECIFIC H.On 13/03/24 04:55, olcott wrote:Since it is a logical impossibility to determine the truthOn 3/12/2024 10:49 PM, Richard Damon wrote:Wrong. You said. yourself. that H1 gets the right answer for D.Not quite. It always gets the wrong answer, but only one of them for each quesiton.They all gets the wrong answer on a whole class of questions
value of a self-contradictory expression the requirement
for H to do this is bogus.
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
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.
You don't even know the definiton of an incorrect question.
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Although there is a mapping from some men to YES/NORight, because that question include a presumption of something not actually present.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.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.
there is no mapping from never unmarried men to YES/NO
thus the question is incorrect for all unmarried men.
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
the question in detail. Making sure to always ignore this key detail
<is> cheating.
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⟩))
The QUESTION is:
Does the machine and input described by this input, Halt when run?
same question at all.
∀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.
above set has no correct answer only because each of these answers
are contradicted by the machine that H is contained within.
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⟩))
*The answer must come from elements of the above set*
gets the wrong answer only because whatever answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
Les messages affichés proviennent d'usenet.