On 3/12/2024 10:33 PM, Richard Damon wrote:...
On 3/12/24 4:56 PM, olcott wrote:
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:Every decider/input pair (referenced in the above set) has a >>>>>>>>> corresponding decider/input pair that only differs by the return >>>>>>>>> value of its decider.
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>> (that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different. >>>>>>>>>
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to actually >>>>>> explain it. The same machine always gives the same return value on the >>>>>> same input.
It has taken me twenty years to translate my intuitions into
words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical
besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their >>>> return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
So they aren't identical.
"Identical except ..." means DIFFERENT.
So you LIE
Not at all. I did not know these details until
On 3/14/2024 7:25 PM, immibis wrote:
On 15/03/24 00:45, olcott wrote:
On 3/14/2024 5:37 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.
Then show me which contradicted answer is correct.
D(D) halts so 1 would be the correct answer.
Strawman deception, that is not a contradicted answer from
the following set:
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⟩))
On 3/14/2024 8:06 PM, Richard Damon wrote:
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:
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:The question posed to Ĥ.H has no correct answer, thus not the >>>>>>>>> same question at all.
On 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:
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:I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
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: >>>>>>>>>>>>>>>>>>>>>>
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
Wrong. You said. yourself. that H1 gets the right answer for D.
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.
Shows you are just a LIAR, as there IS a truth value to the expression
that is the requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>
*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* >>>>>>>>>>>>>>>>>
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. >>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question. >>>>>>>>>>>>>>>>
Nope, common technical term.
Cite a source.
Right, because that question include a presumption of something not
The fact that there DOES exist a mapping Halt(M,d) that maps all TuringThat part is true.
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.
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. >>>>>>>>>>>>>>
actually present.
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. >>>>>>>>>>>>>
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
Except that the mapping requested is about the INPUTS to H, not H itsef.
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.
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? >>>>>>>>>
But it DOES.
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.
I didn't say there was.
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.
No, YOU don't understand that the IS a correct answer, just not the one >>>> that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn was
the right answer.
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*
On 3/13/2024 4:44 AM, Mikko wrote:
On 2024-03-13 03:41:18 +0000, olcott said:Not it is not, it is being mistaken.
On 3/12/2024 10:33 PM, Richard Damon wrote:...
On 3/12/24 4:56 PM, olcott wrote:
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>>>> (that are contained within the above specified set)
only differ by return value will both be wrong on the >>>>>>>>>>>>> same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different.
Every decider/input pair (referenced in the above set) has a >>>>>>>>>>> corresponding decider/input pair that only differs by the return >>>>>>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that >>>>>>>>> their question was incorrect because the opposite answer to the >>>>>>>>> same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to actually >>>>>>>> explain it. The same machine always gives the same return value on the >>>>>>>> same input.
It has taken me twenty years to translate my intuitions into
words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical >>>>>>> besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their
return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
So they aren't identical.
"Identical except ..." means DIFFERENT.
So you LIE
Not at all. I did not know these details until
To claim something as truth without knowing it is to lie.
Lies are telling known falsehoods with the intention to deceive.
On 3/15/2024 5:42 AM, Mikko wrote:
On 2024-03-15 00:56:54 +0000, olcott said:
On 3/14/2024 7:25 PM, immibis wrote:
On 15/03/24 00:45, olcott wrote:
On 3/14/2024 5:37 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.
Then show me which contradicted answer is correct.
D(D) halts so 1 would be the correct answer.
Strawman deception, that is not a contradicted answer from
the following set:
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⟩))
False. That "D(D) halts so 1 would be the correct answer." is true.
It is contradicted by H(D,D) that says 0. Being in "following set",
which is not a set, was not asked.
Unless some H(D,D) aborts the simulation of its input D(D) never stops running. The outermost H(D,D) sees this abort criteria first. If the outermost H(D,D) does not abort its simulation then none of them do. Therefore the outermost H(D,D) is correct to abort its simulation.
When the outermost H(D,D) aborts its simulation then the whole chain
is simulation is completely stopped. Thus proving that this criteria
has been met:
Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
(He has neither reviewed nor agreed to anything else in this paper)
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:
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:
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: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:
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:I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
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: >>>>>>>>>>>>>>>>>>>>>>>>
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
Wrong. You said. yourself. that H1 gets the right >>>>>>>>>>>>>>>>>>>>>> answer for D.
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.
Shows you are just a LIAR, as there IS a truth value >>>>>>>>>>>>>>>>>>>> to the expression that is the requirment for ANY >>>>>>>>>>>>>>>>>>>> SPECIFIC H.
*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*
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.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect >>>>>>>>>>>>>>>>>> question.
Nope, common technical term.
Cite a source.
Right, 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. >>>>>>>>>>>>>>>>
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. >>>>>>>>>>>>>>>
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
Except that the mapping requested is about the INPUTS to >>>>>>>>>>>>>> H, not H itsef.
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.
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?
The question posed to Ĥ.H has no correct answer, thus not the >>>>>>>>>>> same question at all.
But it DOES.
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.
I didn't say there was.
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.
No, YOU don't understand that the IS a correct answer, just not
the one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn
was the right answer.
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*
Is a false claim about a strawman deception really the best you can say?
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.
On 3/15/2024 5:56 AM, Mikko wrote:
On 2024-03-13 14:21:18 +0000, olcott said:Is is never the case that a mistake is a lie.
On 3/13/2024 4:44 AM, Mikko wrote:
On 2024-03-13 03:41:18 +0000, olcott said:Not it is not, it is being mistaken.
On 3/12/2024 10:33 PM, Richard Damon wrote:...
On 3/12/24 4:56 PM, olcott wrote:
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>>>>>> (that are contained within the above specified set) >>>>>>>>>>>>>>> only differ by return value will both be wrong on the >>>>>>>>>>>>>>> same pathological input.
You mean a pair of DIFFERENT machines. Any difference is >>>>>>>>>>>>>> different.
Every decider/input pair (referenced in the above set) has a >>>>>>>>>>>>> corresponding decider/input pair that only differs by the >>>>>>>>>>>>> return
value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return >>>>>>>>>>> value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that >>>>>>>>>>> their question was incorrect because the opposite answer to the >>>>>>>>>>> same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to >>>>>>>>>> actually explain it. The same machine always gives the same >>>>>>>>>> return value on the same input.
It has taken me twenty years to translate my intuitions into >>>>>>>>> words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical >>>>>>>>> besides their return value that cannot decide some property of >>>>>>>>> the same input are being asked the same YES/NO question having >>>>>>>>> no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition >>>>>>>> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides >>>>>>>> their return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
So they aren't identical.
"Identical except ..." means DIFFERENT.
So you LIE
Not at all. I did not know these details until
To claim something as truth without knowing it is to lie.
Lies are telling known falsehoods with the intention to deceive.
Opinions about that vary. But an unintended sin is still a sin.
I have corrected several mistakes in the last two weeks.
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
*This is my biggest mistake*
H ⟨Ĥ⟩ ⟨Ĥ⟩ Cannot wait for its input to complete and then provide
a halt status consistent with the behavior of Ĥ ⟨Ĥ⟩.
Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
(1) H applied ⟨Ĥ⟩ ⟨Ĥ⟩ simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(2) which begins at simulated ⟨Ĥ.q0⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
Shows that H ⟨Ĥ⟩ ⟨Ĥ⟩ sees the abort criteria before Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩.
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:
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:
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:The question posed to Ĥ.H has no correct answer, thus not the >>>>>>>>>>> same question at all.
On 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:
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:I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
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: >>>>>>>>>>>>>>>>>>>>>>>>
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
Wrong. You said. yourself. that H1 gets the right answer for D.
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.
Shows you are just a LIAR, as there IS a truth value to the expression
that is the requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>
*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* >>>>>>>>>>>>>>>>>>>
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.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question. >>>>>>>>>>>>>>>>>>
Nope, common technical term.
Cite a source.
Right, because that question include a presumption of something not
The fact that there DOES exist a mapping Halt(M,d) that maps all TuringThat part is true.
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.
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. >>>>>>>>>>>>>>>>
actually present.
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. >>>>>>>>>>>>>>>
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
Except that the mapping requested is about the INPUTS to H, not H itsef.
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.
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? >>>>>>>>>>>
But it DOES.
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.
I didn't say there was.
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.
No, YOU don't understand that the IS a correct answer, just not the one >>>>>> that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn was >>>> the right answer.
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*
Is a false claim about a strawman deception really the best you can say?
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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