olcott <NoOne@NoWhere.com> writes:

> On 8/12/2021 7:53 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/12/2021 6:00 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>
>>>>> The above is a cut-and-paste of the same corrected and clarified
>>>>> version of Linz that I have been referring to for over a year. I
>>>>> didn't change it.
>>>>
>>>> I can't find any example (in an exchange with me) older than 8 weeks.
>>>> Anyway, I can't correct the impression if you can't point to the post
>>>> that made you think I accepted this bizarre set of symbols.
>>>
>>> This is how Linz does it as shown on his page 319 and my page 21
>>> (He has two start states).
>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥ∞
>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥqn
>> You are right -- it's a re-working of what Linz has. My apologies.
>> But I don't "accept it as true" because it's a universally quantified
>
> This is a universal quantifier: ∀

Did you think I didn't know? On the other hand, I was pretty sure you
would not know about implicit quantification.

>> statement about how a TM behaves for all inputs, and I know some inputs
>> fail. What's more, you don't (yet) claim that it's true either. The
>> only case you claim if true is the one where wM = ⟨Ĥ⟩.
>
> It is a template that stipulates the behavior at certain points in the
> state transitions. Since H only refers to deciders it must always
> reach past the last ⊢* state.

You are confused. Anyway, I don't think unravelling your confusion
matters. Only one case matters, and that is with M = Ĥ (your Ĥ, not
Linz's) and on that case you know we disagree. Your Ĥ does not meet the
specification.

>> So while I accept is states how Ĥ should behave, I don't accept that it
>> true (for all M) and I specifically don't accept that it's true for the
>> one M (and wM) you care about.
>
> Expressions of language that are stipulated to be true are impossibly
> false.
>
> In the following BASIC statement:
> 100 let X = 5
> you are not free to disagree that X equals 5.

Let X be a halt decider. Let X be the largest prime. You are *obliged*
to disagree with these on account of them being meaningless.

--
Ben.

On 8/12/21 8:24 AM, olcott wrote:
> On 8/12/2021 3:36 AM, Malcolm McLean wrote:
>> On Thursday, 12 August 2021 at 06:11:30 UTC+1, Jeff Barnett wrote:
>>> On 8/11/2021 5:40 PM, olcott wrote:
>>>> On 8/11/2021 6:32 PM, Jeff Barnett wrote:
>>>>> On 8/11/2021 5:04 PM, Ben Bacarisse wrote:
>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>
>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER
>>>>>>> WM TO H
>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER
>>>>>>> WM TO H
>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER
>>>>>>> WM TO H
>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER
>>>>>>> WM TO H
>>>>>>
>>>>>> I see you are in "paste the same text" again mode. If you think I can
>>>>>> help in any way, do let me know.
>>>>>
>>>>> That's not nice. Aren't you afraid that he'll develop carpal tunnel
>>>>> syndrome? That along with all of his other deficiencies will surely do
>>>>> him in. Perhaps we can invent shortcuts a la LaTeX macros and font
>>>>> switches to help him cut down the strain. He wont be insulted and will
>>>>> take to it as a pig to mud. Just think, he'll have yet another
>>>>> notation to misuse and abuse.
>>>>
>>>> Ben finally acknowledged the point that I was making:
>>>>
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>> if M applied to wM halts, and
>>>>
>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>> if M applied to wM does not halt
>>>>
>>>> // M refers to the TM of the first wM parameter to Ĥ.qx
>>>>
>>>> My proof requires a whole inference chain that cannot proceed to the
>>>> next point until the current point is accepted as correct.
>>> In addition to carpal tunnel, you are delusional. I get the vague
>>> impression reading the above that you are having a schizoid episode.
>>> There is a complete and total discount between what others say to you
>>> and your replies. I suggest you cut out some of the examples you like
>>> best and show them to your therapist. You do have one, don't you? It may
>>> lead to better, more effective treatment. We are all hoping you get
>>> better but schizoid episodes are hard to suppress. I assume meds have
>>> been prescribed for your condition. Are you taking them? Or are you
>>> refusing medical help just as you refuse to allow Ben to heal your
>>> ignorance. A little trust in your betters will go a long way towards
>>> recovery and a cure. Start now before it's to late. God speed to you.
>>> -
>> YOu might be right, but I think it's just frustration. His execution
>> trace shows
>> what appears to be an infinitely recursive process aborted, and
>> everyone is
>> just ignoring that and focusing on the other reality that his H does not
>> return the result that matches the behaviour of the input when run
>> independently.
>>
>> The attempt to use formal math notation when he doesn't understand it
>> doesn't help. That can create the impression of mental incoherence,
>> but in fact
>> it's not that.
>>
>> Despite a mass of posting, we still don't have a good idea how H works.
>
> _P()
> [00000d02](01)  55          push ebp
> [00000d03](02)  8bec        mov ebp,esp
> [00000d05](03)  8b4508      mov eax,[ebp+08]
> [00000d08](01)  50          push eax       // push 2nd Param
> [00000d09](03)  8b4d08      mov ecx,[ebp+08]
> [00000d0c](01)  51          push ecx       // push 1st Param
> [00000d0d](05)  e870feffff  call 00000b82  // call H
> [00000d12](03)  83c408      add esp,+08
> [00000d15](02)  85c0        test eax,eax
> [00000d17](02)  7402        jz 00000d1b
> [00000d19](02)  ebfe        jmp 00000d19
> [00000d1b](01)  5d          pop ebp
> [00000d1c](01)  c3          ret
> Size in bytes:(0027) [00000d1c]
>
>     machine   stack     stack     machine     assembly
>     address   address   data      code        language
>     ========  ========  ========  =========   =============
> ...[00000d0d][00101829][00000d12] e870feffff  call 00000b82  // call H
>
> Begin Local Halt Decider Simulation at Machine Address:d02
> ...[00000d02][002118f1][002118f5] 55          push ebp
> ...[00000d03][002118f1][002118f5] 8bec        mov ebp,esp
> ...[00000d05][002118f1][002118f5] 8b4508      mov eax,[ebp+08]
> ...[00000d08][002118ed][00000d02] 50          push eax       // push P
> ...[00000d09][002118ed][00000d02] 8b4d08      mov ecx,[ebp+08]
> ...[00000d0c][002118e9][00000d02] 51          push ecx       // push P
> ...[00000d0d][002118e5][00000d12] e870feffff  call 00000b82  // call H
>
> ...[00000d02][0025c319][0025c31d] 55          push ebp
> ...[00000d03][0025c319][0025c31d] 8bec        mov ebp,esp
> ...[00000d05][0025c319][0025c31d] 8b4508      mov eax,[ebp+08]
> ...[00000d08][0025c315][00000d02] 50          push eax       // push P
> ...[00000d09][0025c315][00000d02] 8b4d08      mov ecx,[ebp+08]
> ...[00000d0c][0025c311][00000d02] 51          push ecx       // push P
> ...[00000d0d][0025c30d][00000d12] e870feffff  call 00000b82  // call H
> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>
> If this means that you are telling me that:
> (1) You know the x86 language very well.
>
> (2) Can't begin to understand why P would be stuck in infinitely nested
> simulation while H acts as a pure simulator?

Yes, if H (henceforth Hn) acts like a pure simulator FOREVER, then you
have an infinite loop. The problem is this Hn never gives an answer for
Hn(Hn^,Hn^) so it isn't a correct decider.

Once H know enough to abort its simulation (henceforth Ha) to answer
Ha(Ha^,Ha^) then it needs to use that same Ha for what it analyses Ha^,
and that Ha doesn't create an infinity nested simulation, just one
nested one more level then Ha will simulate, so Ha gets it wrong.

Remember, H^ (aka P) is based on H, so as you postulate different
version of H, each get their OWN H^ they have to get right.

The fact that the aborting Ha can get right the Hm^ for the non-aborting
Hn doesn't prove anything.

As I have summarized before:

Hn^(<Hn^>) is non-Halting
Ha^(<Ha^>) is Halting.

Hn(<Hn^>, <Hn^>) is itself non-Halting so is wrong for not answering
Ha(<Hn^>, <Hn^>) is correct in answering non-Halting, but is uninteresting.

Ha(<Ha^>, <Ha^>( is WRONG in answering non-Halting, as Ha^(<Ha^) is
halting, and THAT is the problem it needs to get right.

It also turns out that

Hn(<Ha^>, <Ha^>) will be correct in answering Halting, which is only
mildly interesting showing that YOUR 'definition' of the correct answer
isn't right based on it being the only answer that can be given.

On 8/12/21 4:36 PM, olcott wrote:
> On 8/12/2021 3:00 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>> if M applied to wM halts, and
>>>
>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>> if M applied to wM does not halt
>>>
>>> // M refers to the TM of the first wM parameter to Ĥ.qx
>>>
>>> Now that you accept that the above is true...
>>
>> I don't.  It's a garbled formula arising from a silly error on your
>> part.  I'd like to know where (you think) I said I accept this nonsense
>> so I can correct any such impression.
>>
>>> ... we can move on to the next point. My proof must proceed exactly
>>> one point at a time an cannot possibly move to the next point until
>>> the current point is fully accepted.
>>
>> It would be simpler if we worked though the reasons you are wrong
>> because there are fewer steps.
>>
>> You've stated that
>>
>>    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>> and you accept (at least you keep quoting) that Linz requires that this
>> should be the case only if Ĥ applied to ⟨Ĥ⟩ does not halt.  QED.
>>
>>> That you believe that the fact that Ĥ applied to ⟨Ĥ⟩ transitions to
>>> its final state of Ĥ.qn and halts nullifies the fact that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
>>> does correctly decide that its input never halts is the next point.
>>
>> ⟨Ĥ⟩ ⟨Ĥ⟩ is a string that encodes a halting computation as shown a few
>> lines above.  That Ĥ applied to ⟨Ĥ⟩ halts does not "nullify" anything,
>> it's just wrong as clearly stated by Linz.
>>
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> ⟨Ĥ⟩ ⟨Ĥ⟩ is not a string that encodes a halting computation.

Why do you say that? Since <H^> <H^> is the same machine as H^ <H^>
which by your above notation would be H^.q0 <H^> which you just showed
ends up halting at H^.qn it is obviously a Halting Computation.

The fact that the partial simulator inside H^ (its copy of H) didn't
reach the halting state doesn't actually imply anything about
non-halting, only the non-halting of an UNCONDITIONAL simulator show that.

>
> Ĥ is a TM that halts only because
> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input never halts.

WRONG. H^ is a TM That HALTS (PERIOD). It halts because the algorithm at
H^.qx, the copy of the algorithm of H, INCORRECTLY decides that the
input it was given wouldn't halt, because is assumes that copies of H
won't abort their simulation when they will.

>
> When you examine this in its x86utm equivalent of H(P,P) there are no
> loopholes that can slip through the cracks because every detail is
> explicitly encoded in the x86 language.

WRONG.

>
> When we examine this as Ĥ applied to ⟨Ĥ⟩ there are millions of pages of
> Turing machine code that cannot be explicitly specified.

Sure they can, at least if your simulator is a real Turing Equivalent.

Now, the fact that H uses the 'address' trick, which can't be mapped
into Turing Machine just shows that your 'equivalents' aren't in the
same relationship as Linz H and H^.

>
> None-the-less the key element of all this is the fact that if we assume
> that the simulating halt decider at Ĥ.qx is simply a UTM then it becomes
> quite obvious that we have an infinite cycle from Ĥ.qx to Ĥ.q0.

Which is a FALSE assumption. A UTM NEVER aborts its simulation, treating
a machine that does as if it was a UTM is UNSOUND logic.

When you make that assumetion, you made an ASS out of U. (we don't fall
for it so the second part of the classical saying doesn't apply).

>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> If we know that we have an infinite cycle then this knowledge all by
> itself proves that the simulating halt decider at Ĥ.qx must abort the
> simulation of its input which proves that this input never halts.
>
>

UNSOUND.

On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>
>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>> perfectly natural.
>>>>
>>>> I can see why you don't want to point out my mistake
>>>> (because there is no mistake).
>>> I keep pointing them out, and you keep wanting to talk about something
>>> else. The mistakes are very simple:
>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>
>> You keep looking at the wrong part of the computation.
>> You keep looking at the wrong part of the computation.
>> You keep looking at the wrong part of the computation.
>> You keep looking at the wrong part of the computation.
>
> I keep looking at the part that shows you are wrong. I'm not interested
> in the other parts.
>

If you wold look at the other part you would see that it does not show
that I am wrong.

>> Yes the first Ĥ halts.
>
> ... when applied to ⟨Ĥ⟩. And Linz says that's wrong.
>
>> No the first ⟨Ĥ⟩ never halts.
>
> ⟨Ĥ⟩ is a string. It does not halt nor does it not halt -- it's just a
> string. Try saying what you mean.
>
>> I keep saying that I have a white cat and you prove that I am wrong by
>> showing that I do not have a black dog.
>
> No. You keep showing the lines that say that your Ĥ does not meet
> Linz's specification (for the one case you care about). You keep

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ does not halt

> showing us the black dog while saying that you have a white cat. I just
> keep pointing to the back dog you keep posting about.
>
>>> (2) You incorrectly claim that ⟨Ĥ⟩ ⟨Ĥ⟩ is a string representing a
>>> computation that does not halt.
>>
>> Ĥ.q0 ⟨Ĥ⟩ copies its input then
>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ simulates this input with this copy
>>
>> which copies its input
>> which is simulated with this copy
>>
>> which copies its input
>> which is simulated with this copy
>>
>> which copies its input
>> which is simulated with this copy...
>
> ⟨Ĥ⟩ ⟨Ĥ⟩ is a string representing the computation that you keep telling us
> halts.
>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 8/12/2021 8:51 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/12/2021 7:53 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/12/2021 6:00 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>>> The above is a cut-and-paste of the same corrected and clarified
>>>>>> version of Linz that I have been referring to for over a year. I
>>>>>> didn't change it.
>>>>>
>>>>> I can't find any example (in an exchange with me) older than 8 weeks.
>>>>> Anyway, I can't correct the impression if you can't point to the post
>>>>> that made you think I accepted this bizarre set of symbols.
>>>>
>>>> This is how Linz does it as shown on his page 319 and my page 21
>>>> (He has two start states).
>>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥ∞
>>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥqn
>>> You are right -- it's a re-working of what Linz has. My apologies.
>>> But I don't "accept it as true" because it's a universally quantified
>>
>> This is a universal quantifier: ∀
>
> Did you think I didn't know? On the other hand, I was pretty sure you
> would not know about implicit quantification.
>
>>> statement about how a TM behaves for all inputs, and I know some inputs
>>> fail. What's more, you don't (yet) claim that it's true either. The
>>> only case you claim if true is the one where wM = ⟨Ĥ⟩.
>>
>> It is a template that stipulates the behavior at certain points in the
>> state transitions. Since H only refers to deciders it must always
>> reach past the last ⊢* state.
>
> You are confused. Anyway, I don't think unravelling your confusion
> matters. Only one case matters, and that is with M = Ĥ (your Ĥ, not
> Linz's) and on that case you know we disagree. Your Ĥ does not meet the
> specification.
>
>>> So while I accept is states how Ĥ should behave, I don't accept that it
>>> true (for all M) and I specifically don't accept that it's true for the
>>> one M (and wM) you care about.
>>
>> Expressions of language that are stipulated to be true are impossibly
>> false.
>>
>> In the following BASIC statement:
>> 100 let X = 5
>> you are not free to disagree that X equals 5.
>
> Let X be a halt decider. Let X be the largest prime. You are *obliged*
> to disagree with these on account of them being meaningless.
>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 8/12/21 10:21 PM, olcott wrote:
> On 8/12/2021 8:51 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/12/2021 7:53 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/12/2021 6:00 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>>> The above is a cut-and-paste of the same corrected and clarified
>>>>>>> version of Linz that I have been referring to for over a year. I
>>>>>>> didn't change it.
>>>>>>
>>>>>> I can't find any example (in an exchange with me) older than 8 weeks.
>>>>>> Anyway, I can't correct the impression if you can't point to the post
>>>>>> that made you think I accepted this bizarre set of symbols.
>>>>>
>>>>> This is how Linz does it as shown on his page 319 and my page 21
>>>>> (He has two start states).
>>>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥ∞
>>>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥqn
>>>> You are right -- it's a re-working of what Linz has.  My apologies.
>>>> But I don't "accept it as true" because it's a universally quantified
>>>
>>> This is a universal quantifier: ∀
>>
>> Did you think I didn't know?  On the other hand, I was pretty sure you
>> would not know about implicit quantification.
>>
>>>> statement about how a TM behaves for all inputs, and I know some inputs
>>>> fail.  What's more, you don't (yet) claim that it's true either.  The
>>>> only case you claim if true is the one where wM = ⟨Ĥ⟩.
>>>
>>> It is a template that stipulates the behavior at certain points in the
>>> state transitions. Since H only refers to deciders it must always
>>> reach past the last ⊢* state.
>>
>> You are confused.  Anyway, I don't think unravelling your confusion
>> matters.  Only one case matters, and that is with M = Ĥ (your Ĥ, not
>> Linz's) and on that case you know we disagree.  Your Ĥ does not meet the
>> specification.
>>
>>>> So while I accept is states how Ĥ should behave, I don't accept that it
>>>> true (for all M) and I specifically don't accept that it's true for the
>>>> one M (and wM) you care about.
>>>
>>> Expressions of language that are stipulated to be true are impossibly
>>> false.
>>>
>>> In the following BASIC statement:
>>> 100 let X = 5
>>> you are not free to disagree that X equals 5.
>>
>> Let X be a halt decider.  Let X be the largest prime.  You are *obliged*
>> to disagree with these on account of them being meaningless.
>>
>
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>
>

No, H^ Halts, PERIOD. Doesn't matter why, it does.

It Halts because the copy of H inside H^ aborted its simulation of its
input <H^> <H^>. 'Inputs' don't halt or not, and aborted simulations do
not imply that the machine being simulated is non-halting.

olcott <NoOne@NoWhere.com> writes:

> On 8/12/2021 8:51 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/12/2021 7:53 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/12/2021 6:00 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>>> The above is a cut-and-paste of the same corrected and clarified
>>>>>>> version of Linz that I have been referring to for over a year. I
>>>>>>> didn't change it.
>>>>>>
>>>>>> I can't find any example (in an exchange with me) older than 8 weeks.
>>>>>> Anyway, I can't correct the impression if you can't point to the post
>>>>>> that made you think I accepted this bizarre set of symbols.
>>>>>
>>>>> This is how Linz does it as shown on his page 319 and my page 21
>>>>> (He has two start states).
>>>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥ∞
>>>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥqn
>>>> You are right -- it's a re-working of what Linz has. My apologies.
>>>> But I don't "accept it as true" because it's a universally quantified
>>>
>>> This is a universal quantifier: ∀
>> Did you think I didn't know? On the other hand, I was pretty sure you
>> would not know about implicit quantification.
>>
>>>> statement about how a TM behaves for all inputs, and I know some inputs
>>>> fail. What's more, you don't (yet) claim that it's true either. The
>>>> only case you claim if true is the one where wM = ⟨Ĥ⟩.
>>>
>>> It is a template that stipulates the behavior at certain points in the
>>> state transitions. Since H only refers to deciders it must always
>>> reach past the last ⊢* state.
>> You are confused. Anyway, I don't think unravelling your confusion
>> matters. Only one case matters, and that is with M = Ĥ (your Ĥ, not
>> Linz's) and on that case you know we disagree. Your Ĥ does not meet the
>> specification.
>>
>>>> So while I accept is states how Ĥ should behave, I don't accept that it
>>>> true (for all M) and I specifically don't accept that it's true for the
>>>> one M (and wM) you care about.
>>>
>>> Expressions of language that are stipulated to be true are impossibly
>>> false.
>>>
>>> In the following BASIC statement:
>>> 100 let X = 5
>>> you are not free to disagree that X equals 5.
>> Let X be a halt decider. Let X be the largest prime. You are *obliged*
>> to disagree with these on account of them being meaningless.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Good to see your 'control' and 'V' keys are working. Anytime you want
to address the points put to you, just let me know.

--
Ben.

On Thursday, 12 August 2021 at 16:50:19 UTC+1, olcott wrote:
> On 8/12/2021 10:40 AM, Malcolm McLean wrote:
> > On Thursday, 12 August 2021 at 13:24:46 UTC+1, olcott wrote:
> >> On 8/12/2021 3:36 AM, Malcolm McLean wrote:
> >>> On Thursday, 12 August 2021 at 06:11:30 UTC+1, Jeff Barnett wrote:
> >>>> On 8/11/2021 5:40 PM, olcott wrote:
> >>>>> On 8/11/2021 6:32 PM, Jeff Barnett wrote:
> >>>>>> On 8/11/2021 5:04 PM, Ben Bacarisse wrote:
> >>>>>>> olcott <No...@NoWhere.com> writes:
> >>>>>>>
> >>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> >>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> >>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> >>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> >>>>>>>
> >>>>>>> I see you are in "paste the same text" again mode. If you think I can
> >>>>>>> help in any way, do let me know.
> >>>>>>
> >>>>>> That's not nice. Aren't you afraid that he'll develop carpal tunnel
> >>>>>> syndrome? That along with all of his other deficiencies will surely do
> >>>>>> him in. Perhaps we can invent shortcuts a la LaTeX macros and font
> >>>>>> switches to help him cut down the strain. He wont be insulted and will
> >>>>>> take to it as a pig to mud. Just think, he'll have yet another
> >>>>>> notation to misuse and abuse.
> >>>>>
> >>>>> Ben finally acknowledged the point that I was making:
> >>>>>
> >>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
> >>>>> if M applied to wM halts, and
> >>>>>
> >>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> >>>>> if M applied to wM does not halt
> >>>>>
> >>>>> // M refers to the TM of the first wM parameter to Ĥ.qx
> >>>>>
> >>>>> My proof requires a whole inference chain that cannot proceed to the
> >>>>> next point until the current point is accepted as correct.
> >>>> In addition to carpal tunnel, you are delusional. I get the vague
> >>>> impression reading the above that you are having a schizoid episode.
> >>>> There is a complete and total discount between what others say to you
> >>>> and your replies. I suggest you cut out some of the examples you like
> >>>> best and show them to your therapist. You do have one, don't you? It may
> >>>> lead to better, more effective treatment. We are all hoping you get
> >>>> better but schizoid episodes are hard to suppress. I assume meds have
> >>>> been prescribed for your condition. Are you taking them? Or are you
> >>>> refusing medical help just as you refuse to allow Ben to heal your
> >>>> ignorance. A little trust in your betters will go a long way towards
> >>>> recovery and a cure. Start now before it's to late. God speed to you..
> >>>> -
> >>> YOu might be right, but I think it's just frustration. His execution trace shows
> >>> what appears to be an infinitely recursive process aborted, and everyone is
> >>> just ignoring that and focusing on the other reality that his H does not
> >>> return the result that matches the behaviour of the input when run independently.
> >>>
> >>> The attempt to use formal math notation when he doesn't understand it
> >>> doesn't help. That can create the impression of mental incoherence, but in fact
> >>> it's not that.
> >>>
> >>> Despite a mass of posting, we still don't have a good idea how H works.
> >> _P()
> >> [00000d02](01) 55 push ebp
> >> [00000d03](02) 8bec mov ebp,esp
> >> [00000d05](03) 8b4508 mov eax,[ebp+08]
> >> [00000d08](01) 50 push eax // push 2nd Param
> >> [00000d09](03) 8b4d08 mov ecx,[ebp+08]
> >> [00000d0c](01) 51 push ecx // push 1st Param
> >> [00000d0d](05) e870feffff call 00000b82 // call H
> >> [00000d12](03) 83c408 add esp,+08
> >> [00000d15](02) 85c0 test eax,eax
> >> [00000d17](02) 7402 jz 00000d1b
> >> [00000d19](02) ebfe jmp 00000d19
> >> [00000d1b](01) 5d pop ebp
> >> [00000d1c](01) c3 ret
> >> Size in bytes:(0027) [00000d1c]
> >> machine stack stack machine assembly
> >> address address data code language
> >> ======== ======== ======== ========= =============
> >> ...[00000d0d][00101829][00000d12] e870feffff call 00000b82 // call H
> >>
> >> Begin Local Halt Decider Simulation at Machine Address:d02
> >> ...[00000d02][002118f1][002118f5] 55 push ebp
> >> ...[00000d03][002118f1][002118f5] 8bec mov ebp,esp
> >> ...[00000d05][002118f1][002118f5] 8b4508 mov eax,[ebp+08]
> >> ...[00000d08][002118ed][00000d02] 50 push eax // push P
> >> ...[00000d09][002118ed][00000d02] 8b4d08 mov ecx,[ebp+08]
> >> ...[00000d0c][002118e9][00000d02] 51 push ecx // push P
> >> ...[00000d0d][002118e5][00000d12] e870feffff call 00000b82 // call H
> >>
> >> ...[00000d02][0025c319][0025c31d] 55 push ebp
> >> ...[00000d03][0025c319][0025c31d] 8bec mov ebp,esp
> >> ...[00000d05][0025c319][0025c31d] 8b4508 mov eax,[ebp+08]
> >> ...[00000d08][0025c315][00000d02] 50 push eax // push P
> >> ...[00000d09][0025c315][00000d02] 8b4d08 mov ecx,[ebp+08]
> >> ...[00000d0c][0025c311][00000d02] 51 push ecx // push P
> >> ...[00000d0d][0025c30d][00000d12] e870feffff call 00000b82 // call H
> >> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
> >> If this means that you are telling me that:
> >> (1) You know the x86 language very well.
> >>
> >> (2) Can't begin to understand why P would be stuck in infinitely nested
> >> simulation while H acts as a pure simulator?
> >>
> > OK, so let's start.
> > There's a line which says "Begin Local Halt Decider Simulation at Machine
> > Address:d02". Is that part of the execution trace, or is it a commented added
> > by hand? If it is part of the execution trace, why doesn't this line appear
> > before the second line with the label 0d02?
> The halt decider automatically writes this when it begins the simulation
> that its halt analysis is based on.
>
OK, so why don't we get a second "Begin Local Halt Decider ..." line output?
> >>
> >> I would say that you must be a liar.
> >>
> > I have a PhD. PhD holder attach a very high value to honesty, in particular
> > with respect to their subjects. But also more generally. I always draw attention
> > if I'm given too much change at the pub, for example, not becaue I'm better
> > than other people, but because I'm a doctor, and a doctor can't be seen to
> > take money that doesn't belong to him.
> >
> So do (1) and (2) apply to you?
>
That was a demand for an apology, but you are too focused on your own
interests to be sensitive to what other people are trying to communicate to you.

olcott <NoOne@NoWhere.com> writes:

> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>
>>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>>> perfectly natural.
>>>>>
>>>>> I can see why you don't want to point out my mistake
>>>>> (because there is no mistake).
>>>> I keep pointing them out, and you keep wanting to talk about something
>>>> else. The mistakes are very simple:
>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>
>>> You keep looking at the wrong part of the computation.
>>> You keep looking at the wrong part of the computation.
>>> You keep looking at the wrong part of the computation.
>>> You keep looking at the wrong part of the computation.
>>
>> I keep looking at the part that shows you are wrong. I'm not interested
>> in the other parts.
>
> If you wold look at the other part you would see that it does not show
> that I am wrong.

That's why I will keep the focus on that parts that do. One day, you
might think it worth addressing these points. I won't re-iterate them
here as the rest of your post just contains the same text that shows
exactly how your Ĥ fails. You obviously can't see it, but I can't see
the point in a cut-and-paste exchange.

If you are puzzled by anything I've written, I'll certainly try to find
a better explanation, but you have to say what it is you are not sure
about.

--
Ben.

On 8/13/2021 3:58 AM, Malcolm McLean wrote:
> On Thursday, 12 August 2021 at 16:50:19 UTC+1, olcott wrote:
>> On 8/12/2021 10:40 AM, Malcolm McLean wrote:
>>> On Thursday, 12 August 2021 at 13:24:46 UTC+1, olcott wrote:
>>>> On 8/12/2021 3:36 AM, Malcolm McLean wrote:
>>>>> On Thursday, 12 August 2021 at 06:11:30 UTC+1, Jeff Barnett wrote:
>>>>>> On 8/11/2021 5:40 PM, olcott wrote:
>>>>>>> On 8/11/2021 6:32 PM, Jeff Barnett wrote:
>>>>>>>> On 8/11/2021 5:04 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>
>>>>>>>>> I see you are in "paste the same text" again mode. If you think I can
>>>>>>>>> help in any way, do let me know.
>>>>>>>>
>>>>>>>> That's not nice. Aren't you afraid that he'll develop carpal tunnel
>>>>>>>> syndrome? That along with all of his other deficiencies will surely do
>>>>>>>> him in. Perhaps we can invent shortcuts a la LaTeX macros and font
>>>>>>>> switches to help him cut down the strain. He wont be insulted and will
>>>>>>>> take to it as a pig to mud. Just think, he'll have yet another
>>>>>>>> notation to misuse and abuse.
>>>>>>>
>>>>>>> Ben finally acknowledged the point that I was making:
>>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>> if M applied to wM halts, and
>>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>> if M applied to wM does not halt
>>>>>>>
>>>>>>> // M refers to the TM of the first wM parameter to Ĥ.qx
>>>>>>>
>>>>>>> My proof requires a whole inference chain that cannot proceed to the
>>>>>>> next point until the current point is accepted as correct.
>>>>>> In addition to carpal tunnel, you are delusional. I get the vague
>>>>>> impression reading the above that you are having a schizoid episode.
>>>>>> There is a complete and total discount between what others say to you
>>>>>> and your replies. I suggest you cut out some of the examples you like
>>>>>> best and show them to your therapist. You do have one, don't you? It may
>>>>>> lead to better, more effective treatment. We are all hoping you get
>>>>>> better but schizoid episodes are hard to suppress. I assume meds have
>>>>>> been prescribed for your condition. Are you taking them? Or are you
>>>>>> refusing medical help just as you refuse to allow Ben to heal your
>>>>>> ignorance. A little trust in your betters will go a long way towards
>>>>>> recovery and a cure. Start now before it's to late. God speed to you.
>>>>>> -
>>>>> YOu might be right, but I think it's just frustration. His execution trace shows
>>>>> what appears to be an infinitely recursive process aborted, and everyone is
>>>>> just ignoring that and focusing on the other reality that his H does not
>>>>> return the result that matches the behaviour of the input when run independently.
>>>>>
>>>>> The attempt to use formal math notation when he doesn't understand it
>>>>> doesn't help. That can create the impression of mental incoherence, but in fact
>>>>> it's not that.
>>>>>
>>>>> Despite a mass of posting, we still don't have a good idea how H works.
>>>> _P()
>>>> [00000d02](01) 55 push ebp
>>>> [00000d03](02) 8bec mov ebp,esp
>>>> [00000d05](03) 8b4508 mov eax,[ebp+08]
>>>> [00000d08](01) 50 push eax // push 2nd Param
>>>> [00000d09](03) 8b4d08 mov ecx,[ebp+08]
>>>> [00000d0c](01) 51 push ecx // push 1st Param
>>>> [00000d0d](05) e870feffff call 00000b82 // call H
>>>> [00000d12](03) 83c408 add esp,+08
>>>> [00000d15](02) 85c0 test eax,eax
>>>> [00000d17](02) 7402 jz 00000d1b
>>>> [00000d19](02) ebfe jmp 00000d19
>>>> [00000d1b](01) 5d pop ebp
>>>> [00000d1c](01) c3 ret
>>>> Size in bytes:(0027) [00000d1c]
>>>> machine stack stack machine assembly
>>>> address address data code language
>>>> ======== ======== ======== ========= =============
>>>> ...[00000d0d][00101829][00000d12] e870feffff call 00000b82 // call H
>>>>
>>>> Begin Local Halt Decider Simulation at Machine Address:d02
>>>> ...[00000d02][002118f1][002118f5] 55 push ebp
>>>> ...[00000d03][002118f1][002118f5] 8bec mov ebp,esp
>>>> ...[00000d05][002118f1][002118f5] 8b4508 mov eax,[ebp+08]
>>>> ...[00000d08][002118ed][00000d02] 50 push eax // push P
>>>> ...[00000d09][002118ed][00000d02] 8b4d08 mov ecx,[ebp+08]
>>>> ...[00000d0c][002118e9][00000d02] 51 push ecx // push P
>>>> ...[00000d0d][002118e5][00000d12] e870feffff call 00000b82 // call H
>>>>
>>>> ...[00000d02][0025c319][0025c31d] 55 push ebp
>>>> ...[00000d03][0025c319][0025c31d] 8bec mov ebp,esp
>>>> ...[00000d05][0025c319][0025c31d] 8b4508 mov eax,[ebp+08]
>>>> ...[00000d08][0025c315][00000d02] 50 push eax // push P
>>>> ...[00000d09][0025c315][00000d02] 8b4d08 mov ecx,[ebp+08]
>>>> ...[00000d0c][0025c311][00000d02] 51 push ecx // push P
>>>> ...[00000d0d][0025c30d][00000d12] e870feffff call 00000b82 // call H
>>>> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>>>> If this means that you are telling me that:
>>>> (1) You know the x86 language very well.
>>>>
>>>> (2) Can't begin to understand why P would be stuck in infinitely nested
>>>> simulation while H acts as a pure simulator?
>>>>
>>> OK, so let's start.
>>> There's a line which says "Begin Local Halt Decider Simulation at Machine
>>> Address:d02". Is that part of the execution trace, or is it a commented added
>>> by hand? If it is part of the execution trace, why doesn't this line appear
>>> before the second line with the label 0d02?
>> The halt decider automatically writes this when it begins the simulation
>> that its halt analysis is based on.
>>
> OK, so why don't we get a second "Begin Local Halt Decider ..." line output?

Begin Local Halt Decider Simulation at Machine Address:d02
....[00000d02][002118f1][002118f5] 55 push ebp
....[00000d03][002118f1][002118f5] 8bec mov ebp,esp
....[00000d05][002118f1][002118f5] 8b4508 mov eax,[ebp+08]
....[00000d08][002118ed][00000d02] 50 push eax // push P
....[00000d09][002118ed][00000d02] 8b4d08 mov ecx,[ebp+08]
....[00000d0c][002118e9][00000d02] 51 push ecx // push P
....[00000d0d][002118e5][00000d12] e870feffff call 00000b82 // call H
....[00000d02][0025c319][0025c31d] 55 push ebp
....[00000d03][0025c319][0025c31d] 8bec mov ebp,esp
....[00000d05][0025c319][0025c31d] 8b4508 mov eax,[ebp+08]
....[00000d08][0025c315][00000d02] 50 push eax // push P
....[00000d09][0025c315][00000d02] 8b4d08 mov ecx,[ebp+08]
....[00000d0c][0025c311][00000d02] 51 push ecx // push P
....[00000d0d][0025c30d][00000d12] e870feffff call 00000b82 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

The simulation of the input begins and continues until it ends.

>>>>
>>>> I would say that you must be a liar.
>>>>
>>> I have a PhD. PhD holder attach a very high value to honesty, in particular
>>> with respect to their subjects. But also more generally. I always draw attention
>>> if I'm given too much change at the pub, for example, not becaue I'm better
>>> than other people, but because I'm a doctor, and a doctor can't be seen to
>>> take money that doesn't belong to him.
>>>
>> So do (1) and (2) apply to you?
>>
> That was a demand for an apology, but you are too focused on your own
> interests to be sensitive to what other people are trying to communicate to you.
>

On 8/13/2021 3:53 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/12/2021 8:51 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/12/2021 7:53 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/12/2021 6:00 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>>> The above is a cut-and-paste of the same corrected and clarified
>>>>>>>> version of Linz that I have been referring to for over a year. I
>>>>>>>> didn't change it.
>>>>>>>
>>>>>>> I can't find any example (in an exchange with me) older than 8 weeks.
>>>>>>> Anyway, I can't correct the impression if you can't point to the post
>>>>>>> that made you think I accepted this bizarre set of symbols.
>>>>>>
>>>>>> This is how Linz does it as shown on his page 319 and my page 21
>>>>>> (He has two start states).
>>>>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥ∞
>>>>>> q0 wM ⊢* Ĥq0 wM wM ⊢* Ĥqn
>>>>> You are right -- it's a re-working of what Linz has. My apologies.
>>>>> But I don't "accept it as true" because it's a universally quantified
>>>>
>>>> This is a universal quantifier: ∀
>>> Did you think I didn't know? On the other hand, I was pretty sure you
>>> would not know about implicit quantification.
>>>
>>>>> statement about how a TM behaves for all inputs, and I know some inputs
>>>>> fail. What's more, you don't (yet) claim that it's true either. The
>>>>> only case you claim if true is the one where wM = ⟨Ĥ⟩.
>>>>
>>>> It is a template that stipulates the behavior at certain points in the
>>>> state transitions. Since H only refers to deciders it must always
>>>> reach past the last ⊢* state.
>>> You are confused. Anyway, I don't think unravelling your confusion
>>> matters. Only one case matters, and that is with M = Ĥ (your Ĥ, not
>>> Linz's) and on that case you know we disagree. Your Ĥ does not meet the
>>> specification.
>>>
>>>>> So while I accept is states how Ĥ should behave, I don't accept that it
>>>>> true (for all M) and I specifically don't accept that it's true for the
>>>>> one M (and wM) you care about.
>>>>
>>>> Expressions of language that are stipulated to be true are impossibly
>>>> false.
>>>>
>>>> In the following BASIC statement:
>>>> 100 let X = 5
>>>> you are not free to disagree that X equals 5.
>>> Let X be a halt decider. Let X be the largest prime. You are *obliged*
>>> to disagree with these on account of them being meaningless.
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> Ĥ only halts because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>
> Good to see your 'control' and 'V' keys are working. Anytime you want
> to address the points put to you, just let me know.
>

When are you going to understand that the points put to me are
misconceptions?

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 8/13/2021 4:01 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>
>>>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>>>> perfectly natural.
>>>>>>
>>>>>> I can see why you don't want to point out my mistake
>>>>>> (because there is no mistake).
>>>>> I keep pointing them out, and you keep wanting to talk about something
>>>>> else. The mistakes are very simple:
>>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>>
>>>> You keep looking at the wrong part of the computation.
>>>> You keep looking at the wrong part of the computation.
>>>> You keep looking at the wrong part of the computation.
>>>> You keep looking at the wrong part of the computation.
>>>
>>> I keep looking at the part that shows you are wrong. I'm not interested
>>> in the other parts.
>>
>> If you would look at the other part you would see that it does not show
>> that I am wrong.
>
> That's why I will keep the focus on that parts that do.

Ah so you openly admit your dishonestly.

> One day, you
> might think it worth addressing these points. I won't re-iterate them
> here as the rest of your post just contains the same text that shows
> exactly how your Ĥ fails. You obviously can't see it, but I can't see
> the point in a cut-and-paste exchange.
>

Until we know that the paradox is not a contradiction we cannot resolve
the paradox.

> If you are puzzled by anything I've written, I'll certainly try to find
> a better explanation, but you have to say what it is you are not sure
> about.
>

No I am not puzzled now that you have admitted that you are a liar.
Let me know when you will address the points that I need address until
then I will stop responding to you.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

olcott <NoOne@NoWhere.com> writes:

> On 8/13/2021 4:01 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>>>
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>
>>>>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>>>>> perfectly natural.
>>>>>>>
>>>>>>> I can see why you don't want to point out my mistake
>>>>>>> (because there is no mistake).
>>>>>> I keep pointing them out, and you keep wanting to talk about something
>>>>>> else. The mistakes are very simple:
>>>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>>>
>>>>> You keep looking at the wrong part of the computation.
>>>>> You keep looking at the wrong part of the computation.
>>>>> You keep looking at the wrong part of the computation.
>>>>> You keep looking at the wrong part of the computation.
>>>>
>>>> I keep looking at the part that shows you are wrong. I'm not interested
>>>> in the other parts.
>>>
>>> If you would look at the other part you would see that it does not show
>>> that I am wrong.
>> That's why I will keep the focus on that parts that do.
>
> Ah so you openly admit your dishonestly.

Wierd. Had you said that the other parts showed you were right, you
might have a case, but simply not showing that you are wrong is no
reason to look at anything. Almost everything every written does not
show that you are wrong.

>> One day, you
>> might think it worth addressing these points. I won't re-iterate them
>> here as the rest of your post just contains the same text that shows
>> exactly how your Ĥ fails. You obviously can't see it, but I can't see
>> the point in a cut-and-paste exchange.
>
> Until we know that the paradox is not a contradiction we cannot
> resolve the paradox.

There is no paradox. There is no contradiction. I wonder how many
times I will have to say this. You could ask if it baffles you, but
diolch yn fawr, you say you won't reply.

>> If you are puzzled by anything I've written, I'll certainly try to find
>> a better explanation, but you have to say what it is you are not sure
>> about.
>
> No I am not puzzled now that you have admitted that you are a liar.

You are a thoroughly unpleasant person.

> Let me know when you will address the points that I need address until
> then I will stop responding to you.

I've had my fingers crossed that you'd stick to your word on this every
time you "threaten" it, but alas, you keep replying to me.

--
Ben.

On 8/13/2021 3:20 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/13/2021 4:01 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>
>>>>>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>>>>>> perfectly natural.
>>>>>>>>
>>>>>>>> I can see why you don't want to point out my mistake
>>>>>>>> (because there is no mistake).
>>>>>>> I keep pointing them out, and you keep wanting to talk about something
>>>>>>> else. The mistakes are very simple:
>>>>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>>>>
>>>>>> You keep looking at the wrong part of the computation.
>>>>>> You keep looking at the wrong part of the computation.
>>>>>> You keep looking at the wrong part of the computation.
>>>>>> You keep looking at the wrong part of the computation.
>>>>>
>>>>> I keep looking at the part that shows you are wrong. I'm not interested
>>>>> in the other parts.
>>>>
>>>> If you would look at the other part you would see that it does not show
>>>> that I am wrong.
>>> That's why I will keep the focus on that parts that do.
>>
>> Ah so you openly admit your dishonestly.
>
> Wierd. Had you said that the other parts showed you were right, you
> might have a case, but simply not showing that you are wrong is no
> reason to look at anything. Almost everything every written does not
> show that you are wrong.
>

While H remains in pure simulation mode simulating the input to H(P,P)
this simulated input never halts thus conclusively proving that H
decides this input correctly.

>>> One day, you
>>> might think it worth addressing these points. I won't re-iterate them
>>> here as the rest of your post just contains the same text that shows
>>> exactly how your Ĥ fails. You obviously can't see it, but I can't see
>>> the point in a cut-and-paste exchange.
>>
>> Until we know that the paradox is not a contradiction we cannot
>> resolve the paradox.
>
> There is no paradox. There is no contradiction. I wonder how many
> times I will have to say this. You could ask if it baffles you, but
> diolch yn fawr, you say you won't reply.
>

Maybe you Are just getting senile in that you say there is no
contradiction.

It is proven that H decides that its input never halts, if nothing
contradicts this then what is your basis for disagreement?

>>> If you are puzzled by anything I've written, I'll certainly try to find
>>> a better explanation, but you have to say what it is you are not sure
>>> about.
>>
>> No I am not puzzled now that you have admitted that you are a liar.
>
> You are a thoroughly unpleasant person.
>

I can be much more pleasant with people that are not liars.

>> Let me know when you will address the points that I need address until
>> then I will stop responding to you.
>
> I've had my fingers crossed that you'd stick to your word on this every
> time you "threaten" it, but alas, you keep replying to me.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

olcott <NoOne@NoWhere.com> writes:

> On 8/13/2021 3:20 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/13/2021 4:01 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>
>>>>>>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>>>>>>> perfectly natural.
>>>>>>>>>
>>>>>>>>> I can see why you don't want to point out my mistake
>>>>>>>>> (because there is no mistake).
>>>>>>>> I keep pointing them out, and you keep wanting to talk about something
>>>>>>>> else. The mistakes are very simple:
>>>>>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>>>>>
>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>
>>>>>> I keep looking at the part that shows you are wrong. I'm not interested
>>>>>> in the other parts.
>>>>>
>>>>> If you would look at the other part you would see that it does not show
>>>>> that I am wrong.
>>>> That's why I will keep the focus on that parts that do.
>>>
>>> Ah so you openly admit your dishonestly.
>> Wierd. Had you said that the other parts showed you were right, you
>> might have a case, but simply not showing that you are wrong is no
>> reason to look at anything. Almost everything every written does not
>> show that you are wrong.
>>
>
> While H remains in pure simulation mode simulating the input to H(P,P) this simulated input never halts thus conclusively proving that H
> decides this input correctly.
>
>>>> One day, you
>>>> might think it worth addressing these points. I won't re-iterate them
>>>> here as the rest of your post just contains the same text that shows
>>>> exactly how your Ĥ fails. You obviously can't see it, but I can't see
>>>> the point in a cut-and-paste exchange.
>>>
>>> Until we know that the paradox is not a contradiction we cannot
>>> resolve the paradox.
>> There is no paradox. There is no contradiction. I wonder how many
>> times I will have to say this. You could ask if it baffles you, but
>> diolch yn fawr, you say you won't reply.
>>
>
> Maybe you Are just getting senile in that you say there is no contradiction.
>
> It is proven that H decides that its input never halts, if nothing contradicts this then what is your basis for disagreement?
>
>>>> If you are puzzled by anything I've written, I'll certainly try to find
>>>> a better explanation, but you have to say what it is you are not sure
>>>> about.
>>>
>>> No I am not puzzled now that you have admitted that you are a liar.
>> You are a thoroughly unpleasant person.
>>
>
> I can be much more pleasant with people that are not liars.
>
>>> Let me know when you will address the points that I need address until
>>> then I will stop responding to you.
>> I've had my fingers crossed that you'd stick to your word on this every
>> time you "threaten" it, but alas, you keep replying to me.

Why did you reply?

--
Ben.

On 8/13/2021 4:28 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/13/2021 3:20 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/13/2021 4:01 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>
>>>>>>>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>>>>>>>> perfectly natural.
>>>>>>>>>>
>>>>>>>>>> I can see why you don't want to point out my mistake
>>>>>>>>>> (because there is no mistake).
>>>>>>>>> I keep pointing them out, and you keep wanting to talk about something
>>>>>>>>> else. The mistakes are very simple:
>>>>>>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>>>>>>
>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>
>>>>>>> I keep looking at the part that shows you are wrong. I'm not interested
>>>>>>> in the other parts.
>>>>>>
>>>>>> If you would look at the other part you would see that it does not show
>>>>>> that I am wrong.
>>>>> That's why I will keep the focus on that parts that do.
>>>>
>>>> Ah so you openly admit your dishonestly.
>>> Wierd. Had you said that the other parts showed you were right, you
>>> might have a case, but simply not showing that you are wrong is no
>>> reason to look at anything. Almost everything every written does not
>>> show that you are wrong.
>>>
>>
>> While H remains in pure simulation mode simulating the input to H(P,P) this simulated input never halts thus conclusively proving that H
>> decides this input correctly.
>>
>>>>> One day, you
>>>>> might think it worth addressing these points. I won't re-iterate them
>>>>> here as the rest of your post just contains the same text that shows
>>>>> exactly how your Ĥ fails. You obviously can't see it, but I can't see
>>>>> the point in a cut-and-paste exchange.
>>>>
>>>> Until we know that the paradox is not a contradiction we cannot
>>>> resolve the paradox.
>>> There is no paradox. There is no contradiction. I wonder how many
>>> times I will have to say this. You could ask if it baffles you, but
>>> diolch yn fawr, you say you won't reply.
>>>
>>
>> Maybe you Are just getting senile in that you say there is no contradiction.
>>
>> It is proven that H decides that its input never halts, if nothing contradicts this then what is your basis for disagreement?
>>
>>>>> If you are puzzled by anything I've written, I'll certainly try to find
>>>>> a better explanation, but you have to say what it is you are not sure
>>>>> about.
>>>>
>>>> No I am not puzzled now that you have admitted that you are a liar.
>>> You are a thoroughly unpleasant person.
>>>
>>
>> I can be much more pleasant with people that are not liars.
>>
>>>> Let me know when you will address the points that I need address until
>>>> then I will stop responding to you.
>>> I've had my fingers crossed that you'd stick to your word on this every
>>> time you "threaten" it, but alas, you keep replying to me.
>
> Why did you reply?
>

Why will you not give this an honest review?

While H remains in pure simulation mode simulating the input to H(P,P)
this simulated input never halts thus conclusively proving that H
decides this input correctly.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

olcott <NoOne@NoWhere.com> writes:

> On 8/13/2021 4:28 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/13/2021 3:20 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/13/2021 4:01 AM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>>
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>>
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>>
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>>
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>
>>>>>>>>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>>>>>>>>> perfectly natural.
>>>>>>>>>>>
>>>>>>>>>>> I can see why you don't want to point out my mistake
>>>>>>>>>>> (because there is no mistake).
>>>>>>>>>> I keep pointing them out, and you keep wanting to talk about something
>>>>>>>>>> else. The mistakes are very simple:
>>>>>>>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>>>>>>>
>>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>>
>>>>>>>> I keep looking at the part that shows you are wrong. I'm not interested
>>>>>>>> in the other parts.
>>>>>>>
>>>>>>> If you would look at the other part you would see that it does not show
>>>>>>> that I am wrong.
>>>>>> That's why I will keep the focus on that parts that do.
>>>>>
>>>>> Ah so you openly admit your dishonestly.
>>>> Wierd. Had you said that the other parts showed you were right, you
>>>> might have a case, but simply not showing that you are wrong is no
>>>> reason to look at anything. Almost everything every written does not
>>>> show that you are wrong.
>>>>
>>>
>>> While H remains in pure simulation mode simulating the input to H(P,P) this simulated input never halts thus conclusively proving that H
>>> decides this input correctly.
>>>
>>>>>> One day, you
>>>>>> might think it worth addressing these points. I won't re-iterate them
>>>>>> here as the rest of your post just contains the same text that shows
>>>>>> exactly how your Ĥ fails. You obviously can't see it, but I can't see
>>>>>> the point in a cut-and-paste exchange.
>>>>>
>>>>> Until we know that the paradox is not a contradiction we cannot
>>>>> resolve the paradox.
>>>> There is no paradox. There is no contradiction. I wonder how many
>>>> times I will have to say this. You could ask if it baffles you, but
>>>> diolch yn fawr, you say you won't reply.
>>>>
>>>
>>> Maybe you Are just getting senile in that you say there is no contradiction.
>>>
>>> It is proven that H decides that its input never halts, if nothing contradicts this then what is your basis for disagreement?
>>>
>>>>>> If you are puzzled by anything I've written, I'll certainly try to find
>>>>>> a better explanation, but you have to say what it is you are not sure
>>>>>> about.
>>>>>
>>>>> No I am not puzzled now that you have admitted that you are a liar.
>>>> You are a thoroughly unpleasant person.
>>>>
>>>
>>> I can be much more pleasant with people that are not liars.
>>>
>>>>> Let me know when you will address the points that I need address until
>>>>> then I will stop responding to you.
>>>> I've had my fingers crossed that you'd stick to your word on this every
>>>> time you "threaten" it, but alas, you keep replying to me.
>> Why did you reply?
>
> Why will you not give this an honest review?

Unless it is to apologise, you could keep to your word and stop
responding to me.

--
Ben.

On 8/13/2021 5:06 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/13/2021 4:28 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/13/2021 3:20 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/13/2021 4:01 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>>>
>>>>>>>>>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>>>>>>>>>> perfectly natural.
>>>>>>>>>>>>
>>>>>>>>>>>> I can see why you don't want to point out my mistake
>>>>>>>>>>>> (because there is no mistake).
>>>>>>>>>>> I keep pointing them out, and you keep wanting to talk about something
>>>>>>>>>>> else. The mistakes are very simple:
>>>>>>>>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>>>>>>>>
>>>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>>>>
>>>>>>>>> I keep looking at the part that shows you are wrong. I'm not interested
>>>>>>>>> in the other parts.
>>>>>>>>
>>>>>>>> If you would look at the other part you would see that it does not show
>>>>>>>> that I am wrong.
>>>>>>> That's why I will keep the focus on that parts that do.
>>>>>>
>>>>>> Ah so you openly admit your dishonestly.
>>>>> Wierd. Had you said that the other parts showed you were right, you
>>>>> might have a case, but simply not showing that you are wrong is no
>>>>> reason to look at anything. Almost everything every written does not
>>>>> show that you are wrong.
>>>>>
>>>>
>>>> While H remains in pure simulation mode simulating the input to H(P,P) this simulated input never halts thus conclusively proving that H
>>>> decides this input correctly.
>>>>
>>>>>>> One day, you
>>>>>>> might think it worth addressing these points. I won't re-iterate them
>>>>>>> here as the rest of your post just contains the same text that shows
>>>>>>> exactly how your Ĥ fails. You obviously can't see it, but I can't see
>>>>>>> the point in a cut-and-paste exchange.
>>>>>>
>>>>>> Until we know that the paradox is not a contradiction we cannot
>>>>>> resolve the paradox.
>>>>> There is no paradox. There is no contradiction. I wonder how many
>>>>> times I will have to say this. You could ask if it baffles you, but
>>>>> diolch yn fawr, you say you won't reply.
>>>>>
>>>>
>>>> Maybe you Are just getting senile in that you say there is no contradiction.
>>>>
>>>> It is proven that H decides that its input never halts, if nothing contradicts this then what is your basis for disagreement?
>>>>
>>>>>>> If you are puzzled by anything I've written, I'll certainly try to find
>>>>>>> a better explanation, but you have to say what it is you are not sure
>>>>>>> about.
>>>>>>
>>>>>> No I am not puzzled now that you have admitted that you are a liar.
>>>>> You are a thoroughly unpleasant person.
>>>>>
>>>>
>>>> I can be much more pleasant with people that are not liars.
>>>>
>>>>>> Let me know when you will address the points that I need address until
>>>>>> then I will stop responding to you.
>>>>> I've had my fingers crossed that you'd stick to your word on this every
>>>>> time you "threaten" it, but alas, you keep replying to me.
>>> Why did you reply?
>>
>> Why will you not give this an honest review?
>
> Unless it is to apologise, you could keep to your word and stop
> responding to me.
>

I changed my mind I will instead repeat this endlessly until I get a
fair an honest review.

Why will you not give this an honest review?

While H remains in pure simulation mode simulating the input to H(P,P)
this simulated input never halts thus conclusively proving that H
decides this input correctly.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 8/13/2021 4:01 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>
>>>>>>> I can see you don't want to talk about why you are wrong. That's
>>>>>>> perfectly natural.
>>>>>>
>>>>>> I can see why you don't want to point out my mistake
>>>>>> (because there is no mistake).
>>>>> I keep pointing them out, and you keep wanting to talk about something
>>>>> else. The mistakes are very simple:
>>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>>
>>>> You keep looking at the wrong part of the computation.
>>>> You keep looking at the wrong part of the computation.
>>>> You keep looking at the wrong part of the computation.
>>>> You keep looking at the wrong part of the computation.
>>>
>>> I keep looking at the part that shows you are wrong. I'm not interested
>>> in the other parts.
>>
>> If you wold look at the other part you would see that it does not show
>> that I am wrong.
>
> That's why I will keep the focus on that parts that do.

I see no possible correct way to interpret this as other than
dishonestly. You utterly refuse to look at the part that proves that I
am correct while maintaining the view that I am incorrect.

Why will you not give this an honest review?

While H remains in pure simulation mode simulating the input to H(P,P)
this simulated input never halts thus conclusively proving that H
decides this input correctly.

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

olcott <NoOne@NoWhere.com> writes:

> On 8/13/2021 5:06 PM, Ben Bacarisse wrote:

>> Unless it is to apologise, you could keep to your word and stop
>> responding to me.
>
> I changed my mind

I doubt anyone is surprised.

> I will instead repeat this endlessly until I get a fair an honest
> review.

Presumably in a bid to avoid ever publishing either a paper of the key
bit of code you are hiding (or, more likely, don't have). That's fine
by me.

More then 30 months after claiming to have a TM that everyone told you
was impossible, you are reduced to begging for comments from someone you
think is a liar. That's an odd use of your time.

--
Ben.

On 8/13/2021 5:25 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/13/2021 5:06 PM, Ben Bacarisse wrote:
>
>>> Unless it is to apologise, you could keep to your word and stop
>>> responding to me.
>>
>> I changed my mind
>
> I doubt anyone is surprised.
>
>> I will instead repeat this endlessly until I get a fair an honest
>> review.
>
> Presumably in a bid to avoid ever publishing either a paper of the key
> bit of code you are hiding (or, more likely, don't have). That's fine
> by me.
>
> More then 30 months after claiming to have a TM that everyone told you
> was impossible, you are reduced to begging for comments from someone you
> think is a liar. That's an odd use of your time.
>

_P()
[00000d02](01) 55 push ebp
[00000d03](02) 8bec mov ebp,esp
[00000d05](03) 8b4508 mov eax,[ebp+08]
[00000d08](01) 50 push eax // push 2nd Param
[00000d09](03) 8b4d08 mov ecx,[ebp+08]
[00000d0c](01) 51 push ecx // push 1st Param
[00000d0d](05) e870feffff call 00000b82 // call H
[00000d12](03) 83c408 add esp,+08
[00000d15](02) 85c0 test eax,eax
[00000d17](02) 7402 jz 00000d1b
[00000d19](02) ebfe jmp 00000d19
[00000d1b](01) 5d pop ebp
[00000d1c](01) c3 ret
Size in bytes:(0027) [00000d1c]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
....[00000d0d][00101829][00000d12] e870feffff call 00000b82 // call H

Begin Local Halt Decider Simulation at Machine Address:d02
....[00000d02][002118f1][002118f5] 55 push ebp
....[00000d03][002118f1][002118f5] 8bec mov ebp,esp
....[00000d05][002118f1][002118f5] 8b4508 mov eax,[ebp+08]
....[00000d08][002118ed][00000d02] 50 push eax // push P
....[00000d09][002118ed][00000d02] 8b4d08 mov ecx,[ebp+08]
....[00000d0c][002118e9][00000d02] 51 push ecx // push P
....[00000d0d][002118e5][00000d12] e870feffff call 00000b82 // call H
....[00000d02][0025c319][0025c31d] 55 push ebp
....[00000d03][0025c319][0025c31d] 8bec mov ebp,esp
....[00000d05][0025c319][0025c31d] 8b4508 mov eax,[ebp+08]
....[00000d08][0025c315][00000d02] 50 push eax // push P
....[00000d09][0025c315][00000d02] 8b4d08 mov ecx,[ebp+08]
....[00000d0c][0025c311][00000d02] 51 push ecx // push P
....[00000d0d][0025c30d][00000d12] e870feffff call 00000b82 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

It is very obvious that while H remains in pure simulator mode that the
simulated P cannot possibly stop running. The first 7 instructions of P
must endlessly repeat.

Before I can submit my paper for publication I must make my words clear
enough that they are obviously correct.

If my above words were unclear or incorrect someone would be able to
point out the unclarity or error.

The fact that everyone changes the subject instead of pointing out any
error or unclarity seems to have no reasonable assessment other than
their dishonesty.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On Friday, August 13, 2021 at 3:47:39 PM UTC-7, olcott wrote:
>
> Before I can submit my paper for publication I must make my words clear
> enough that they are obviously correct.
>
> If my above words were unclear or incorrect someone would be able to
> point out the unclarity or error.
>
> The fact that everyone changes the subject instead of pointing out any
> error or unclarity seems to have no reasonable assessment other than
> their dishonesty.
If you are serious lets start over. First let us agree on exactly what you
going to prove. Make what you consider your answer to the question
"What in this paper going to prove?" and we will opine whether the
statement is clear and accurate. Remember this is you writing to
convince others that what you have to say is worth reading.

In a new thread, of course.

On 8/13/21 4:56 PM, olcott wrote:
> On 8/13/2021 3:20 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/13/2021 4:01 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/12/2021 8:49 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/12/2021 8:05 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/12/2021 7:26 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 8/12/2021 6:28 PM, olcott wrote:
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> Can you see that the above never halts if Ĥ.qx is a UTM?
>>>>>>>>>>
>>>>>>>>>> I can see you don't want to talk about why you are wrong.  That's
>>>>>>>>>> perfectly natural.
>>>>>>>>>
>>>>>>>>> I can see why you don't want to point out my mistake
>>>>>>>>> (because there is no mistake).
>>>>>>>> I keep pointing them out, and you keep wanting to talk about
>>>>>>>> something
>>>>>>>> else.  The mistakes are very simple:
>>>>>>>> (1) Ĥ applied to ⟨Ĥ⟩ halts when Linz says it should not.
>>>>>>>
>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>> You keep looking at the wrong part of the computation.
>>>>>>
>>>>>> I keep looking at the part that shows you are wrong.  I'm not
>>>>>> interested
>>>>>> in the other parts.
>>>>>
>>>>> If you would look at the other part you would see that it does not
>>>>> show
>>>>> that I am wrong.
>>>> That's why I will keep the focus on that parts that do.
>>>
>>> Ah so you openly admit your dishonestly.
>>
>> Wierd.  Had you said that the other parts showed you were right, you
>> might have a case, but simply not showing that you are wrong is no
>> reason to look at anything.  Almost everything every written does not
>> show that you are wrong.
>>
>
> While H remains in pure simulation mode simulating the input to H(P,P)
> this simulated input never halts thus conclusively proving that H
> decides this input correctly.

Except that H then DOES abort its simulation, and thus proves it never
was a pure simulator, so any logic based on that assumption is UNSOUND.

Remember, a pure simulator is one that NEVER aborts its simulation.
N-E-V-E-R.

It will, by definition, simulate a non-halting input for an unbounded
number of steps.

A simulator that stops simulating after a finite bounded number of steps
fails to have ever been a pure simulator.

You are just using a wrong definition, and thus get UNSOUND results.

Since you have a FALSE premise early in your proof, your WHOLE proof is
thus shown to be UNSOUND.

Until your provide some suitable refutation, we can all just say UNSOUND
every time you make this statement.

On 8/13/2021 6:54 PM, dklei...@gmail.com wrote:
> On Friday, August 13, 2021 at 3:47:39 PM UTC-7, olcott wrote:
>>
>> Before I can submit my paper for publication I must make my words clear
>> enough that they are obviously correct.
>>
>> If my above words were unclear or incorrect someone would be able to
>> point out the unclarity or error.
>>
>> The fact that everyone changes the subject instead of pointing out any
>> error or unclarity seems to have no reasonable assessment other than
>> their dishonesty.
>
> If you are serious lets start over. First let us agree on exactly what you
> going to prove. Make what you consider your answer to the question
> "What in this paper going to prove?" and we will opine whether the
> statement is clear and accurate. Remember this is you writing to
> convince others that what you have to say is worth reading.
>
> In a new thread, of course.
>

That sounds like a great idea.

I will only seek to prove that H(P,P) decides its input correctly.
This means that any rebuttal that does not pertain to H actually
deciding its actual input would be off topic.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 8/13/2021 6:54 PM, dklei...@gmail.com wrote:
> On Friday, August 13, 2021 at 3:47:39 PM UTC-7, olcott wrote:
>>
>> Before I can submit my paper for publication I must make my words clear
>> enough that they are obviously correct.
>>
>> If my above words were unclear or incorrect someone would be able to
>> point out the unclarity or error.
>>
>> The fact that everyone changes the subject instead of pointing out any
>> error or unclarity seems to have no reasonable assessment other than
>> their dishonesty.
>
> If you are serious lets start over. First let us agree on exactly what you
> going to prove. Make what you consider your answer to the question
> "What in this paper going to prove?" and we will opine whether the
> statement is clear and accurate. Remember this is you writing to
> convince others that what you have to say is worth reading.
>
> In a new thread, of course.
>

I did this several times with this one as the best one:
[How do we know H(P,P)==0 is the correct halt status for the input to H?
(HTML for fixed width font)]

With the fixed width font it is much easier to study the x86 code
execution trace having fixed width column headings.

I also changed the order of the material so that it more closely was in
prerequisite order.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein