On 8/5/2021 11:41 AM, Jeff Barnett wrote:
> On 8/5/2021 10:08 AM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>
>>> For instance, .999999999999999 is close to 1, however, its not
>>> arbitrarily close to 1 like .999... is
>>
>> The reals have the property that if x =/= y there are reals between x
>> and y.  If 0.999... is only "arbitrarily close to 1" what reals lie
>> between it an 1?  0.999... == 1.
>
> Recently, I've been trying to remember the name of that property.  I
> think I once saw something where a set R was (totally) ordered by < with
> the "in between" property, one could say that "R was dense in <". Even
> if that phrasing is correct, I'd like to find a better sounding and more
> familiar term for the property. Do you recall such a name or have any
> suggestions?

Whops! I think that may be the other way round: "< is dense in R". Not
really sure and can't recall where I say this (in either form).
--
Jeff Barnett

On 05/08/2021 18:41, Jeff Barnett wrote:
> On 8/5/2021 10:08 AM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>
>>> For instance, .999999999999999 is close to 1, however, its not
>>> arbitrarily close to 1 like .999... is
>>
>> The reals have the property that if x =/= y there are reals between x
>> and y.  If 0.999... is only "arbitrarily close to 1" what reals lie
>> between it an 1?  0.999... == 1.
>
> Recently, I've been trying to remember the name of that property.  I
> think I once saw something where a set R was (totally) ordered by < with
> the "in between" property, one could say that "R was dense in <". Even
> if that phrasing is correct, I'd like to find a better sounding and more
> familiar term for the property. Do you recall such a name or have any
> suggestions?

I'd say there are a couple of distinct (related) ideas, one topological
and one order-based.

For the topology idea, an example:
The rationals are dense in the reals, because given any real number
there are rational numbers arbitrarily close. (Or put another way every
real number is the limit of some sequence of rationals. Or the closure
of the set of rationals (within the reals) is the full set of reals.

The topology idea isn't using orderings, and there's another idea called
"dense ordering" which sounds more exactly like what you said - the
ordering < is a dense ordering on the reals because given a < b there is
always a c (in the reals) with a < c < b. I guess we might also say
that the rationals are order-dense in the reals because c can be chosen
to be rational... not totally sure if that's normal to say or not.

If you're happy with just real numbers, i.e. allowing c above to be real
rather than rational, then of course if a < b then

a < (a+b)/2 < b

so c = (a+b)/2 = (the average of a and b) always works.)

Also, a bit different, there's the Archimedean property of the reals
which says that given two positive reals a,b, we can always exceed b by
adding together sufficiently many copies of a. This is the basic
property that implies that between any two real numbers there is a
rational number, and also effectively rules out infinitesimal and
infinite quantities within the reals. [That doesn't mean the reals
can't be extended to a larger system featuring such quantities, but then
we have something other than the real numbers of everyday mathematics.]

Some links:
<https://en.wikipedia.org/wiki/Dense_set>
<https://en.wikipedia.org/wiki/Dense_order>
<https://en.wikipedia.org/wiki/Archimedean_property>

Mike.

On 8/5/2021 9:08 AM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>
>> For instance, .999999999999999 is close to 1, however, its not
>> arbitrarily close to 1 like .999... is
>
> The reals have the property that if x =/= y there are reals between x
> and y. If 0.999... is only "arbitrarily close to 1" what reals lie
> between it an 1? 0.999... == 1.
>

Exactly.

On 8/5/2021 12:30 PM, Mike Terry wrote:
> On 05/08/2021 18:41, Jeff Barnett wrote:
>> On 8/5/2021 10:08 AM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> For instance, .999999999999999 is close to 1, however, its not
>>>> arbitrarily close to 1 like .999... is
>>>
>>> The reals have the property that if x =/= y there are reals between x
>>> and y.  If 0.999... is only "arbitrarily close to 1" what reals lie
>>> between it an 1?  0.999... == 1.
>>
>> Recently, I've been trying to remember the name of that property.  I
>> think I once saw something where a set R was (totally) ordered by <
>> with the "in between" property, one could say that "R was dense in <".
>> Even if that phrasing is correct, I'd like to find a better sounding
>> and more familiar term for the property. Do you recall such a name or
>> have any suggestions?
>
>
> I'd say there are a couple of distinct (related) ideas, one topological
> and one order-based.
>
> For the topology idea, an example:
> The rationals are dense in the reals, because given any real number
> there are rational numbers arbitrarily close.  (Or put another way every
> real number is the limit of some sequence of rationals.  Or the closure
> of the set of rationals (within the reals) is the full set of reals.
>
> The topology idea isn't using orderings, and there's another idea called
> "dense ordering" which sounds more exactly like what you said - the
> ordering < is a dense ordering on the reals because given a < b there is
> always a c (in the reals) with a < c < b.  I guess we might also say
> that the rationals are order-dense in the reals because c can be chosen
> to be rational... not totally sure if that's normal to say or not.
>
> If you're happy with just real numbers, i.e. allowing c above to be real
> rather than rational, then of course if a < b then
>
>   a < (a+b)/2 < b
>
> so c = (a+b)/2 = (the average of a and b) always works.)
>
> Also, a bit different, there's the Archimedean property of the reals
> which says that given two positive reals a,b, we can always exceed b by
> adding together sufficiently many copies of a.  This is the basic
> property that implies that between any two real numbers there is a
> rational number, and also effectively rules out infinitesimal and
> infinite quantities within the reals.  [That doesn't mean the reals
> can't be extended to a larger system featuring such quantities, but then
> we have something other than the real numbers of everyday mathematics.]
>
> Some links:
>    <https://en.wikipedia.org/wiki/Dense_set>
>    <https://en.wikipedia.org/wiki/Dense_order>
>    <https://en.wikipedia.org/wiki/Archimedean_property>

This makes me think of using convergents of continued fractions. Using
infinite convergents gets us arbitrarily close to the target real.

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

> On 8/5/2021 9:08 AM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>
>>> For instance, .999999999999999 is close to 1, however, its not
>>> arbitrarily close to 1 like .999... is
>> The reals have the property that if x =/= y there are reals between x
>> and y. If 0.999... is only "arbitrarily close to 1" what reals lie
>> between it an 1? 0.999... == 1.
>
> Exactly.

I must have misunderstood. I thought you were saying that .999... is
arbitrarily close to 1, which implies, to me, that it's not 1.

--
Ben.

On 8/5/2021 1:30 PM, Mike Terry wrote:
> On 05/08/2021 18:41, Jeff Barnett wrote:
>> On 8/5/2021 10:08 AM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> For instance, .999999999999999 is close to 1, however, its not
>>>> arbitrarily close to 1 like .999... is
>>>
>>> The reals have the property that if x =/= y there are reals between x
>>> and y.  If 0.999... is only "arbitrarily close to 1" what reals lie
>>> between it an 1?  0.999... == 1.
>>
>> Recently, I've been trying to remember the name of that property.  I
>> think I once saw something where a set R was (totally) ordered by <
>> with the "in between" property, one could say that "R was dense in <".
>> Even if that phrasing is correct, I'd like to find a better sounding
>> and more familiar term for the property. Do you recall such a name or
>> have any suggestions?
>
>
> I'd say there are a couple of distinct (related) ideas, one topological
> and one order-based.
>
> For the topology idea, an example:
> The rationals are dense in the reals, because given any real number
> there are rational numbers arbitrarily close.  (Or put another way every
> real number is the limit of some sequence of rationals.  Or the closure
> of the set of rationals (within the reals) is the full set of reals.
>
> The topology idea isn't using orderings, and there's another idea called
> "dense ordering" which sounds more exactly like what you said - the
> ordering < is a dense ordering on the reals because given a < b there is
> always a c (in the reals) with a < c < b.  I guess we might also say
> that the rationals are order-dense in the reals because c can be chosen
> to be rational... not totally sure if that's normal to say or not.
>
> If you're happy with just real numbers, i.e. allowing c above to be real
> rather than rational, then of course if a < b then
>
>   a < (a+b)/2 < b
>
> so c = (a+b)/2 = (the average of a and b) always works.)
>
> Also, a bit different, there's the Archimedean property of the reals
> which says that given two positive reals a,b, we can always exceed b by
> adding together sufficiently many copies of a.  This is the basic
> property that implies that between any two real numbers there is a
> rational number, and also effectively rules out infinitesimal and
> infinite quantities within the reals.  [That doesn't mean the reals
> can't be extended to a larger system featuring such quantities, but then
> we have something other than the real numbers of everyday mathematics.]
>
> Some links:
>    <https://en.wikipedia.org/wiki/Dense_set>
>    <https://en.wikipedia.org/wiki/Dense_order>
>    <https://en.wikipedia.org/wiki/Archimedean_property>

I have a set that is strictly linearly order by >: That means > is anti
symmetric, anti reflexive, and obeys transitivity and trichotomy. I then
want to say that if for every a > b, there is a c such that a>c>b. To
introduce "Archimedean" into the discuss drags in arithmetic and I don't
want or need that baggage. So I'll probably say that "> is a dense
ordering on X" then explain what I meant. Thanks for the reply.
--
Jeff Barnett

olcott <NoOne@NoWhere.com> writes:

> On 8/4/2021 6:23 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/4/2021 1:51 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/1/2021 11:00 AM, Ben Bacarisse wrote:

>>>>>> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn. This clearly
>>>>>> shows that Ĥ applied to ⟨Ĥ⟩ halts. You can see the final state right
>>>>>
>>>>> 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
>>>>>
>>>>> You are using the wrong Ĥ.
>>>> First of all, let's be 100% clear: I am talking about what /your/ Ĥ
>>>> does, based in the facts you have let slip about it.
>>>>
>>>>> Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this
>>>>> ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.
>>>>
>>>> No. How many years have you been staring at this one page from Linz?
>>>> You still don't know what it says. Do ask me questions, if you'd like
>>>> to know what the text you've been sure is wrong for 17 years really
>>>> says.
>>>>
>>>
>>> ...Turing machine halting problem.
>>> Simply stated, the problem is:
>>> given the description of a Turing machine M
>>> given the description of a Turing machine M
>>> given the description of a Turing machine M
>>> given the description of a Turing machine M
>>> given the description of a Turing machine M
>>>
>>> and an input w, does M, when started in the initial configuration q0w, perform a computation that eventually halts?
>>>
>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>
>>> Ĥ.q0 ⟨M⟩ ⊢* Ĥ.qx ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qn
>>> if M applied to ⟨M⟩ does not halt
>>>
>>> When ⟨M⟩ = ⟨Ĥ⟩: I have proved that Ĥ ⟨M⟩ transitions to Ĥ.qn because M
>>> never reaches a final state.
>>
>> This a schoolboy error. Why are you so scared to ask me questions? Do
>> you fear you might understand me?
>
> Speaking to me with denigration debases only yourself.

"You can't hold a coherent thought for even a fraction of a second"
"There is no halt decider mentioned here numbskull"
"If you weren't dumber than a box of rocks..."
"The halting problem is theory of computation material, nitwit"

and so on and so on. I could find dozens more from you.

> The halting problem is not about deciding whether or not a Turing
> Machine halts, it is only about whether or not the description of a
> Turing machine specifies a computation that reaches its final state.

No. Mind you, the error here is just in the writing rather than the
logic. If you could say what you mean, it might be correct.

You are right that the halting problem is not about deciding whether or
not a Turing machine halts because it's about computations and not
Turing machines. But its not about whether or not the description of a
Turing machine specifies a computation that reaches its final state
because the description of a Turing machine does not specify a
computation at all.

Anyway, you've not addressed the schoolboy error. If (when ⟨M⟩ = ⟨Ĥ⟩)
Ĥ ⟨M⟩ transitions to Ĥ.qn then M (AKA Ĥ) reaches its final state. It's
just daft to say that you can prove that Ĥ transitions to its final
state (Ĥ.qn) because it (Ĥ) never reaches a final state!

> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Next time, remember to add the text from Linz that shows you are wrong:
"if Ĥ applied to ŵ halts" for the first and "if Ĥ applied to ŵ does not
halt" for the second. (ŵ is what Linz calls Ĥ.)

--
Ben.

olcott <NoOne@NoWhere.com> writes:

> On 8/4/2021 5:23 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/4/2021 1:37 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:

>>>>>> You are wrong because
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> where Linz requires that
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>
>>>>> You have the above incorrectly in these two
>>>>> M refers to the Turing Machine described by wM.
>>>>>
>>>>> 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
>>>>>
>>>>> This is *NOT* the Ĥ that is executed.
>>>>
>>>> I have no idea what you are objecting to. You can substitute Ĥ and ⟨Ĥ⟩
>>>> into what Linz says, can't you? I did it above and you don't like it:
>>>> For M = Ĥ and wM = ⟨Ĥ⟩ Linz tells us that
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>
>>>> Your Ĥ is not like that. Its behaviour is boring and obvious and
>>>> trivial.
>>>
>>> This is how it really is:
>>> if ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>>> and ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its final state.
>> No. ⟨Ĥ⟩ is a string, not a TM. The TM that string represents is Ĥ and
>> Ĥ applied to ⟨Ĥ⟩ does halt.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> The question is not: Does Ĥ halt on its input?

Yes it is. You deliberately missed out the words used by Linz for each
of those cases: "if Ĥ applied to ŵ halts", and "if Ĥ applied to ŵ does
not halt" (ŵ is his name for ⟨Ĥ⟩). As you can see this is /exactly/ the
question that matters.

Mind you, it's better to specify the input (as Linz does) rather than to
say "its input". The context (Ĥ.q0 ⟨Ĥ⟩) makes it clear in this case,
but there are lots of times when I have no idea what "its" refers to.
You do it with "its simulation" too.

> The question is: Does the Turing Machine description ⟨Ĥ⟩ specify a
> computation that reaches its final state on input ⟨Ĥ⟩?
>
> Do you agree with this?

No. ⟨Ĥ⟩ does not specify a computation.

--
Ben.

olcott <NoOne@NoWhere.com> writes:

> On 8/4/2021 6:22 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/4/2021 2:22 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/4/2021 7:53 AM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/2/2021 8:45 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>>>>> As long as it is simply dismissed out-of-hand as a contradiction the
>>>>>>>>> paradox remains unresolved.
>>>>>>>>
>>>>>>>> There is no contradiction or paradox. You Ĥ is just the wrong sort of
>>>>>>>> TM. The proof you want to "refute" is talking about this sort of Ĥ:
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>>
>>>>>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>>>>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>>>>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>>>>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>>>>>
>>>>>> Maybe saying it a couple more times will help. After four times I can
>>>>>> tell you that it's still wrong. Maybe about a dozen more?
>>>>>>
>>>>>> Whether what happens after Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correct or not is determined
>>>>>> by Linz, not by you. And you are clear that
>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn.
>>>>>
>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>> if M applied to wM does not halt
>>>>>
>>>>> As explained in complete detail below:
>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> Yes, please don't tell me the final state yet again. This is not been
>>>> in dispute for some time.
>>>>
>>>>> because M applied to wM does not halt
>>>>> where M is Machine_of(⟨Ĥ⟩) (1st param) above
>>>>> and wM is ⟨Ĥ⟩ the second param above.
>>>>>
>>>>> Because wM is referring to ⟨Ĥ⟩ and M is referring to the underlying
>>>>> machine of ⟨Ĥ⟩ the last line above is translated to: if
>>>>> Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ does not halt
>>>> That's convoluted. ⟨Ĥ⟩ is the encoding of Ĥ so to find out what Linz
>>>> expects from Ĥ applied to ⟨Ĥ⟩ we just substitute M = Ĥ and wM = ⟨Ĥ⟩ into
>>>> the above:
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>
>>> It is not the first Ĥ that is being referred to it is *only* the
>>> machine represented by the input ⟨Ĥ⟩ that is being referred to.
>> The machine represented by ⟨Ĥ⟩ is Ĥ. There is only one Ĥ being
>> discussed here -- yours.
>>
>>> That machine
>>> never reaches its final state.
>> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn, or so you've told us. Are you changing
>> your story? I'd like an answer. It's not a hard question.
>
> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn

OK, so you are not changing your story. Linz tells us that

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.

so your Ĥ is not interesting. Such TMs are ten-a-penny and none of them
have anything to say about the proof in Linz. You should apologise for
hooking people in with a lie 30 months ago. You don't have anything
that anyone would consider to be impossible. What is impossible, and
you once false claimed to have, is a TM that behaves as Linz says, if
only for this one case.

> because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.

The reason is not relevant. That's good because the above is nonsense,
iteral nonsense. What is the input "to"? It can't be Ĥ.qx because
that's a state and states don't have input. It can't be Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
because that's a TM configuration (state+tape contents) and these don't
have input either.

But it does not matter what this jumble of words means because Ĥ.q0 ⟨Ĥ⟩
transitions to Ĥ.qn and it shouldn't.

>>> Ĥ.q0 ⟨Ĥ[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
>>> if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.
>>>
>>> That you persistently ignore this distinction really seems to be a
>>> diverge from an honest dialogue.
>>
>> If the various strings you've chosen to number (badly) are not identical
>> then your "hat" construction is wrong. Linz's "hat" version makes an
>> exact copy.
>
> When Bill says that his identical twin brother is not going to go to
> the store, and then Bill goes to the store this does not make him a
> liar.

Evasion. Is ⟨Ĥ[i]⟩ = ⟨Ĥ⟩ for all i or not? You need to start answering
simple technical questions like this. If "yes", the notation is there
only to help your exposition, if "no" your Ĥ is not behaving at it
should.

(Twins, unlike strings, are not mathematical objects, but if were to
model twins mathematically we'd have to used non-equal objects since
they can't be substituted one for the other.)

>> You are free to write ⟨Ĥ[99]⟩ if you like, but I am also free to change
>> that back to ⟨Ĥ⟩ because they are identical strings. Anything true of
>
> Ĥ is not a string it is a Turing machine.
> Turing machines are not identical to strings.
> Ĥ halts the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its final state
> whether or not the halt decider at Ĥ.qx stop simulating it.

There some words missing there I think. But it's irrelevant waffle.
The undisputed fact is that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn but Linz's Ĥ
should not behave like this. Everything that happens on the way to qn
is irrelevant.

> If you want an actual honest dialogue we must have closure on some of
> the points. You must say what things you agree with and not merely
> ignore those things that you agree with.

Likewise. But your task is simpler than mine. We both agree that
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn. All you have to agree with is that this is not the
behaviour specified by Linz: Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn if (and only if) Ĥ applied
to ⟨Ĥ⟩ does not halt.

On the other hand, I have to put pick waffle about input to something
that has no input. Strings never halting or not reaching some state
when string are just strings. All of which concerns the details of how
your Ĥ is wrong.

> Do you understand that the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its
> final state whether or not the halt decider at Ĥ.qx stop simulating
> it?

No. I can't parse the sentence. To what does "its" refer? There is
only one final state in Ĥ and it's Ĥ's final state!

There's a deep problem here. You've never written a UTM (I'm not using
"poetic license" here, I mean a UTM) so you don't know how a UTM
simulation proceeds. The string is not decoded into a TM with its own
states. It can't even use the tape in the usual way because the tape is
now used for more than one purpose. Students get this (eventually)
because they read the textbook and do the exercises.

But the part you keep ignoring is why should anyone care, since Ĥ, and
the H it's build from, do not behave as Linz describes?

> If you disagree then to prove that you are not simply being
> disagreeable you must proceed with a dialogue on this single point
> until we reach mutual agreement. All other points will not be
> discussed until we reach mutual agreement on this point.

Nonsense. You've stated and agreed to the fact that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn.
The details of that wrong journey are just something you want me to get
bogged down in. You could simply agree with the plain fact that anyone
could write an Ĥ that behaves like yours and none of them would say
anything about the proof. You'd be free to go a walk some dogs.

--
Ben.

On 05/08/2021 22:51, Jeff Barnett wrote:
> On 8/5/2021 1:30 PM, Mike Terry wrote:
>> On 05/08/2021 18:41, Jeff Barnett wrote:
>>> On 8/5/2021 10:08 AM, Ben Bacarisse wrote:
>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>
>>>>> For instance, .999999999999999 is close to 1, however, its not
>>>>> arbitrarily close to 1 like .999... is
>>>>
>>>> The reals have the property that if x =/= y there are reals between x
>>>> and y.  If 0.999... is only "arbitrarily close to 1" what reals lie
>>>> between it an 1?  0.999... == 1.
>>>
>>> Recently, I've been trying to remember the name of that property.  I
>>> think I once saw something where a set R was (totally) ordered by <
>>> with the "in between" property, one could say that "R was dense in
>>> <". Even if that phrasing is correct, I'd like to find a better
>>> sounding and more familiar term for the property. Do you recall such
>>> a name or have any suggestions?
>>
>>
>> I'd say there are a couple of distinct (related) ideas, one
>> topological and one order-based.
>>
>> For the topology idea, an example:
>> The rationals are dense in the reals, because given any real number
>> there are rational numbers arbitrarily close.  (Or put another way
>> every real number is the limit of some sequence of rationals.  Or the
>> closure of the set of rationals (within the reals) is the full set of
>> reals.
>>
>> The topology idea isn't using orderings, and there's another idea
>> called "dense ordering" which sounds more exactly like what you said -
>> the ordering < is a dense ordering on the reals because given a < b
>> there is always a c (in the reals) with a < c < b.  I guess we might
>> also say that the rationals are order-dense in the reals because c can
>> be chosen to be rational... not totally sure if that's normal to say
>> or not.
>>
>> If you're happy with just real numbers, i.e. allowing c above to be
>> real rather than rational, then of course if a < b then
>>
>>    a < (a+b)/2 < b
>>
>> so c = (a+b)/2 = (the average of a and b) always works.)
>>
>> Also, a bit different, there's the Archimedean property of the reals
>> which says that given two positive reals a,b, we can always exceed b
>> by adding together sufficiently many copies of a.  This is the basic
>> property that implies that between any two real numbers there is a
>> rational number, and also effectively rules out infinitesimal and
>> infinite quantities within the reals.  [That doesn't mean the reals
>> can't be extended to a larger system featuring such quantities, but
>> then we have something other than the real numbers of everyday
>> mathematics.]
>>
>> Some links:
>>     <https://en.wikipedia.org/wiki/Dense_set>
>>     <https://en.wikipedia.org/wiki/Dense_order>
>>     <https://en.wikipedia.org/wiki/Archimedean_property>
>
> I have a set that is strictly linearly order by >: That means > is anti
> symmetric, anti reflexive, and obeys transitivity and trichotomy. I then
> want to say that if for every a > b, there is a c such that a>c>b. To
> introduce "Archimedean" into the discuss drags in arithmetic and I don't
> want or need that baggage. So I'll probably say that "> is a dense
> ordering on X" then explain what I meant. Thanks for the reply.

Yeah, that would be exactly right.
Mike.

On 8/5/2021 2:21 PM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>
>> On 8/5/2021 9:08 AM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> For instance, .999999999999999 is close to 1, however, its not
>>>> arbitrarily close to 1 like .999... is
>>> The reals have the property that if x =/= y there are reals between x
>>> and y. If 0.999... is only "arbitrarily close to 1" what reals lie
>>> between it an 1? 0.999... == 1.
>>
>> Exactly.
>
> I must have misunderstood. I thought you were saying that .999... is
> arbitrarily close to 1, which implies, to me, that it's not 1.
>

Well, imvvho, there may be two ways to look at it. One is that the limit
is, as-is. Okay. Fine. Then the other way is to try, on a step-by-step
basis, think iteration and/or recursion to try to reach the limit. The
limit is 1, fine. However. using our finite brains to meticulously try
to walk on in, means we will never reach the limit because 1 - our
current iterate is a always greater than 1. Therefore,

1 - 1/x*(x - 1) = 0 if we take the infinite arbitrarily close case into
account and apply a limit. However, 1 - 1/x*(x - 1) != 0 if us finite
beings try to compute every fraction, on a step-by-step basis... ;^)

1/1*0, 1/10*9, 1/100*99, 1/1000*999, 1/10000*9999... gets arbitrarily
closer to one... :^)

On 8/5/2021 3:42 PM, Chris M. Thomasson wrote:
> On 8/5/2021 2:21 PM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>
>>> On 8/5/2021 9:08 AM, Ben Bacarisse wrote:
>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>
>>>>> For instance, .999999999999999 is close to 1, however, its not
>>>>> arbitrarily close to 1 like .999... is
>>>> The reals have the property that if x =/= y there are reals between x
>>>> and y.  If 0.999... is only "arbitrarily close to 1" what reals lie
>>>> between it an 1?  0.999... == 1.
>>>
>>> Exactly.
>>
>> I must have misunderstood.  I thought you were saying that .999... is
>> arbitrarily close to 1, which implies, to me, that it's not 1.
>>
>
> Well, imvvho, there may be two ways to look at it. One is that the limit
> is, as-is. Okay. Fine. Then the other way is to try, on a step-by-step
> basis, think iteration and/or recursion to try to reach the limit. The
> limit is 1, fine. However. using our finite brains to meticulously try
> to walk on in, means we will never reach the limit because 1 - our
> current iterate is a always greater than 1[...]

ARGH, I meant to say 1 minus the current iterate is always greater than
zero! Sorry! DOH!!!! Damn it.

On 8/5/2021 3:42 PM, Chris M. Thomasson wrote:
> On 8/5/2021 2:21 PM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>
>>> On 8/5/2021 9:08 AM, Ben Bacarisse wrote:
>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>
>>>>> For instance, .999999999999999 is close to 1, however, its not
>>>>> arbitrarily close to 1 like .999... is
>>>> The reals have the property that if x =/= y there are reals between x
>>>> and y.  If 0.999... is only "arbitrarily close to 1" what reals lie
>>>> between it an 1?  0.999... == 1.
>>>
>>> Exactly.
>>
>> I must have misunderstood.  I thought you were saying that .999... is
>> arbitrarily close to 1, which implies, to me, that it's not 1.
>>
>
> Well, imvvho, there may be two ways to look at it. One is that the limit
> is, as-is. Okay. Fine. Then the other way is to try, on a step-by-step
> basis, think iteration and/or recursion to try to reach the limit. The
> limit is 1, fine. However. using our finite brains to meticulously try
> to walk on in, means we will never reach the limit because 1 - our
> current iterate is a always greater than 1. Therefore,
>
> 1 - 1/x*(x - 1) = 0 if we take the infinite arbitrarily close case into
> account and apply a limit. However, 1 - 1/x*(x - 1) != 0 if us finite
> beings try to compute every fraction, on a step-by-step basis... ;^)
>
> 1/1*0, 1/10*9, 1/100*99, 1/1000*999, 1/10000*9999... gets arbitrarily
> closer to one... :^)

Oh ouch! 1/1*0, Well, shit fire == GI issues!

On 8/5/2021 3:49 PM, Chris M. Thomasson wrote:
> On 8/5/2021 3:42 PM, Chris M. Thomasson wrote:
>> On 8/5/2021 2:21 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 8/5/2021 9:08 AM, Ben Bacarisse wrote:
>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>
>>>>>> For instance, .999999999999999 is close to 1, however, its not
>>>>>> arbitrarily close to 1 like .999... is
>>>>> The reals have the property that if x =/= y there are reals between x
>>>>> and y.  If 0.999... is only "arbitrarily close to 1" what reals lie
>>>>> between it an 1?  0.999... == 1.
>>>>
>>>> Exactly.
>>>
>>> I must have misunderstood.  I thought you were saying that .999... is
>>> arbitrarily close to 1, which implies, to me, that it's not 1.
>>>
>>
>> Well, imvvho, there may be two ways to look at it. One is that the
>> limit is, as-is. Okay. Fine. Then the other way is to try, on a
>> step-by-step basis, think iteration and/or recursion to try to reach
>> the limit. The limit is 1, fine. However. using our finite brains to
>> meticulously try to walk on in, means we will never reach the limit
>> because 1 - our current iterate is a always greater than 1. Therefore,
>>
>> 1 - 1/x*(x - 1) = 0 if we take the infinite arbitrarily close case
>> into account and apply a limit. However, 1 - 1/x*(x - 1) != 0 if us
>> finite beings try to compute every fraction, on a step-by-step
>> basis... ;^)
>>
>> 1/1*0, 1/10*9, 1/100*99, 1/1000*999, 1/10000*9999... gets arbitrarily
>> closer to one... :^)
>
> Oh ouch! 1/1*0, Well, shit fire == GI issues!

The funny part is that over on sci.math, there are some posters that are
convinced that zero does not exist.

On 8/5/2021 5:12 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/4/2021 6:23 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/4/2021 1:51 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
>
>>>>>>> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn. This clearly
>>>>>>> shows that Ĥ applied to ⟨Ĥ⟩ halts. You can see the final state right
>>>>>>
>>>>>> 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
>>>>>>
>>>>>> You are using the wrong Ĥ.
>>>>> First of all, let's be 100% clear: I am talking about what /your/ Ĥ
>>>>> does, based in the facts you have let slip about it.
>>>>>
>>>>>> Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this
>>>>>> ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.
>>>>>
>>>>> No. How many years have you been staring at this one page from Linz?
>>>>> You still don't know what it says. Do ask me questions, if you'd like
>>>>> to know what the text you've been sure is wrong for 17 years really
>>>>> says.
>>>>>
>>>>
>>>> ...Turing machine halting problem.
>>>> Simply stated, the problem is:
>>>> given the description of a Turing machine M
>>>> given the description of a Turing machine M
>>>> given the description of a Turing machine M
>>>> given the description of a Turing machine M
>>>> given the description of a Turing machine M
>>>>
>>>> and an input w, does M, when started in the initial configuration q0w, perform a computation that eventually halts?
>>>>
>>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>>
>>>> Ĥ.q0 ⟨M⟩ ⊢* Ĥ.qx ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qn
>>>> if M applied to ⟨M⟩ does not halt
>>>>
>>>> When ⟨M⟩ = ⟨Ĥ⟩: I have proved that Ĥ ⟨M⟩ transitions to Ĥ.qn because M
>>>> never reaches a final state.
>>>
>>> This a schoolboy error. Why are you so scared to ask me questions? Do
>>> you fear you might understand me?
>>
>> Speaking to me with denigration debases only yourself.
>
> "You can't hold a coherent thought for even a fraction of a second"
> "There is no halt decider mentioned here numbskull"
> "If you weren't dumber than a box of rocks..."
> "The halting problem is theory of computation material, nitwit"
>
> and so on and so on. I could find dozens more from you.
>
>> The halting problem is not about deciding whether or not a Turing
>> Machine halts, it is only about whether or not the description of a
>> Turing machine specifies a computation that reaches its final state.
>
> No. Mind you, the error here is just in the writing rather than the
> logic. If you could say what you mean, it might be correct.
>
> You are right that the halting problem is not about deciding whether or
> not a Turing machine halts because it's about computations and not
> Turing machines. But its not about whether or not the description of a
> Turing machine specifies a computation that reaches its final state
> because the description of a Turing machine does not specify a
> computation at all.
>

the Turing machine halting problem. Simply stated, the problem is: given
the description of a Turing machine M and an input w, does M, when
started in the initial configuration q0w, perform a computation that
eventually halts? U sing an abbreviated way of talking about the
problem, we ask whether M applied to w, or simply (M, w), halts or does
not halt. http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

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 Turing machine of the inputs not the Turing machine that
is being executed.

if Machine_Of(wM) applied to wM does not halt

So when Bill says that his identical twin brother is never going to the
store and then Bill goes to the store THIS IS NOT A CONTRADICTION.

When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn THIS IS NOT A CONTRADICTION.

If you don't want an honest dialogue then please say so.

> Anyway, you've not addressed the schoolboy error. If (when ⟨M⟩ = ⟨Ĥ⟩)
> Ĥ ⟨M⟩ transitions to Ĥ.qn then M (AKA Ĥ) reaches its final state. It's
> just daft to say that you can prove that Ĥ transitions to its final
> state (Ĥ.qn) because it (Ĥ) never reaches a final state!
>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> Next time, remember to add the text from Linz that shows you are wrong:
> "if Ĥ applied to ŵ halts" for the first and "if Ĥ applied to ŵ does not
> halt" for the second. (ŵ is what Linz calls Ĥ.)
>

--
Copyright 2021 Pete Olcott

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

On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/4/2021 5:23 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/4/2021 1:37 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
>
>>>>>>> You are wrong because
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> where Linz requires that
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>
>>>>>> You have the above incorrectly in these two
>>>>>> M refers to the Turing Machine described by wM.
>>>>>>
>>>>>> 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
>>>>>>
>>>>>> This is *NOT* the Ĥ that is executed.
>>>>>
>>>>> I have no idea what you are objecting to. You can substitute Ĥ and ⟨Ĥ⟩
>>>>> into what Linz says, can't you? I did it above and you don't like it:
>>>>> For M = Ĥ and wM = ⟨Ĥ⟩ Linz tells us that
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>
>>>>> Your Ĥ is not like that. Its behaviour is boring and obvious and
>>>>> trivial.
>>>>
>>>> This is how it really is:
>>>> if ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>>>> and ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its final state.
>>> No. ⟨Ĥ⟩ is a string, not a TM. The TM that string represents is Ĥ and
>>> Ĥ applied to ⟨Ĥ⟩ does halt.
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>> The question is not: Does Ĥ halt on its input?
>
> Yes it is.

The question is:
Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
The ansswer to this question is provably no!

Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION

> You deliberately missed out the words used by Linz for each
> of those cases: "if Ĥ applied to ŵ halts", and "if Ĥ applied to ŵ does
> not halt" (ŵ is his name for ⟨Ĥ⟩). As you can see this is /exactly/ the
> question that matters.
>
> Mind you, it's better to specify the input (as Linz does) rather than to
> say "its input". The context (Ĥ.q0 ⟨Ĥ⟩) makes it clear in this case,
> but there are lots of times when I have no idea what "its" refers to.
> You do it with "its simulation" too.
>
>> The question is: Does the Turing Machine description ⟨Ĥ⟩ specify a
>> computation that reaches its final state on input ⟨Ĥ⟩?
>>
>> Do you agree with this?
>
> No. ⟨Ĥ⟩ does not specify a computation.
>

--
Copyright 2021 Pete Olcott

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

On 8/5/21 8:17 PM, olcott wrote:
> On 8/5/2021 5:12 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/4/2021 6:23 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/4/2021 1:51 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
>>
>>>>>>>> We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn.  This
>>>>>>>> clearly
>>>>>>>> shows that Ĥ applied to ⟨Ĥ⟩ halts.  You can see the final state
>>>>>>>> right
>>>>>>>
>>>>>>> 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
>>>>>>>
>>>>>>> You are using the wrong Ĥ.
>>>>>> First of all, let's be 100% clear: I am talking about what /your/ Ĥ
>>>>>> does, based in the facts you have let slip about it.
>>>>>>
>>>>>>> Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of
>>>>>>> this
>>>>>>> ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.
>>>>>>
>>>>>> No.  How many years have you been staring at this one page from Linz?
>>>>>> You still don't know what it says.  Do ask me questions, if you'd
>>>>>> like
>>>>>> to know what the text you've been sure is wrong for 17 years really
>>>>>> says.
>>>>>>
>>>>>
>>>>> ...Turing machine halting problem.
>>>>> Simply stated, the problem is:
>>>>> given the description of a Turing machine M
>>>>> given the description of a Turing machine M
>>>>> given the description of a Turing machine M
>>>>> given the description of a Turing machine M
>>>>> given the description of a Turing machine M
>>>>>
>>>>> and an input w, does M, when started in the initial configuration
>>>>> q0w, perform a computation that eventually halts?
>>>>>
>>>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>>>
>>>>> Ĥ.q0 ⟨M⟩ ⊢* Ĥ.qx ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qn
>>>>> if M applied to ⟨M⟩ does not halt
>>>>>
>>>>> When ⟨M⟩ = ⟨Ĥ⟩: I have proved that Ĥ ⟨M⟩ transitions to Ĥ.qn because M
>>>>> never reaches a final state.
>>>>
>>>> This a schoolboy error.  Why are you so scared to ask me questions?  Do
>>>> you fear you might understand me?
>>>
>>> Speaking to me with denigration debases only yourself.
>>
>>    "You can't hold a coherent thought for even a fraction of a second"
>>    "There is no halt decider mentioned here numbskull"
>>    "If you weren't dumber than a box of rocks..."
>>    "The halting problem is theory of computation material, nitwit"
>>
>> and so on and so on.  I could find dozens more from you.
>>
>>> The halting problem is not about deciding whether or not a Turing
>>> Machine halts, it is only about whether or not the description of a
>>> Turing machine specifies a computation that reaches its final state.
>>
>> No.  Mind you, the error here is just in the writing rather than the
>> logic.  If you could say what you mean, it might be correct.
>>
>> You are right that the halting problem is not about deciding whether or
>> not a Turing machine halts because it's about computations and not
>> Turing machines.  But its not about whether or not the description of a
>> Turing machine specifies a computation that reaches its final state
>> because the description of a Turing machine does not specify a
>> computation at all.
>>
>
> the Turing machine halting problem. Simply stated, the problem is: given
> the description of a Turing machine M and an input w, does M, when
> started in the initial configuration q0w, perform a computation that
> eventually halts? U sing an abbreviated way of talking about the
> problem, we ask whether M applied to w, or simply (M, w), halts or does
> not halt. http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>
>
> 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 Turing machine of the inputs not the Turing machine that
> is being executed.
>

NO. M referes to the Turing machine DESCRIBED by the inputs. The input
is NOT a Turing Machine, but a description, read what you wrote as a
definition and actually USE it.

The notation used does NOT distinctly indicate 'descriptions', which
might make thing cleared, but assumes you are smart enough to realize
that to have a 'machine' as in input implies that you actually give it a
description.

Not only that, but w is ALSO a description, as it is quite possible that
the machie H uses a different alphabet for its tape then the machine M,
so you need to 'encode' what M would get into the 'character set' that H
understands.

> if Machine_Of(wM) applied to wM does not halt
>
> So when Bill says that his identical twin brother is never going to the
> store and then Bill goes to the store THIS IS NOT A CONTRADICTION.

And doesn't apply to Turing Machines.

>
> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn   THIS IS NOT A CONTRADICTION.

So H^(<H^>) is a Halting computation for ALL copies of H^, include the
representation given to H, so H is wrong.

>
> If you don't want an honest dialogue then please say so.

You Too.

>
>> Anyway, you've not addressed the schoolboy error.  If (when ⟨M⟩ = ⟨Ĥ⟩)
>> Ĥ ⟨M⟩ transitions to Ĥ.qn then M (AKA Ĥ) reaches its final state.  It's
>> just daft to say that you can prove that Ĥ transitions to its final
>> state (Ĥ.qn) because it (Ĥ) never reaches a final state!
>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>> Next time, remember to add the text from Linz that shows you are wrong:
>> "if Ĥ applied to ŵ halts" for the first and "if Ĥ applied to ŵ does not
>> halt" for the second.  (ŵ is what Linz calls Ĥ.)
>>
>
>

On 8/5/21 7:07 AM, olcott wrote:
> On 8/5/2021 4:02 AM, Malcolm McLean wrote:
>> On Thursday, 5 August 2021 at 04:18:21 UTC+1, olcott wrote:
>>> On 8/4/2021 6:22 PM, Ben Bacarisse wrote:
>>>> olcott <No...@NoWhere.com> writes:
>>>>
>>>> If the various strings you've chosen to number (badly) are not
>>>> identical
>>>> then your "hat" construction is wrong. Linz's "hat" version makes an
>>>> exact copy.
>>>>
>>> When Bill says that his identical twin brother is not going to go to the
>>> store, and then Bill goes to the store this does not make him a liar.
>>>
>>>> You are free to write ⟨Ĥ[99]⟩ if you like, but I am also free to change
>>>> that back to ⟨Ĥ⟩ because they are identical strings. Anything true of
>>>>
>>> Ĥ is not a string it is a Turing machine.
>>> Turing machines are not identical to strings.
>>> Ĥ halts the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its final state
>>> whether or not the halt decider at Ĥ.qx stop simulating it.
>>>
>>> If you want an actual honest dialogue we must have closure on some of
>>> the points. You must say what things you agree with and not merely
>>> ignore those things that you agree with.
>>>
>>> Do you understand that the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its
>>> final state whether or not the halt decider at Ĥ.qx stop simulating it?
>>>
>> H_Hat <H_Hat> halts. So if it is simulated by a UTM, the simulation
>> UTM <H_HAT><H_Hat> also halts.
>
> It is very easily proven that ⟨Ĥ⟩ on ⟨Ĥ⟩ cannot possibly ever reach its
> final state. If it stops running without reaching is final state then
> this does not count as halting.

Actually that is a MEANINGLESS statement as representations don't execute.

Since you admit that H^(<H^>) does halt, the UTM(<H^>, <H^>) also halts,
and thus the fact that H's simulation is aborted before it gets there
proves nothing except that you don't know what you are talking about.

>
>> But the halt decider embedded in H_Hat is not a UTM. It's a near UTM,
>> that has abort logic. Somewhere the abort logic must be triggered,
>> which causes H_Hat<H_Hat> to halt.
>
> If it stops running without reaching is final state then this does not
> count as halting. It can not possibly reach its final state whether or
> not its simulation is ever aborted, therefore it never halts.

WRONG. If you modify THAT copy (so you don't touch what is on the tape)
we see that H'(<H^>, <H^>) where the H' version has had its aborting
removed (and thus converted into a UTM, while H^ still uses the original
version of H) we see that this will halt and correctly decide that
H^(<H^>) is halting.

This shows that H is wrong.

>
>> H<H_Hat><H_Hat> also triggers this abort logic, so H<H_Hat><H_Hat>
>> reports "false" (non-halting).
>>>
>>> If you disagree then to prove that you are not simply being disagreeable
>>> you must proceed with a dialogue on this single point until we reach
>>> mutual agreement. All other points will not be discussed until we reach
>>> mutual agreement on this point.
>>>
>> We agree that H_Hat<H_Hat> halts.
>> We agree that H <H_Hat> <H_Hat> reports "false" (non-halting).
>>
>> Can't you see the obvious?
>>
>
> The mistake that you are making is counting stopping without reaching a
> final state as halting, it is not.
>

No. H^(<H^>) will ALWAYS run to completion if its execution is not
incorrectly Halted.

On 8/5/21 8:21 PM, olcott wrote:
> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/4/2021 5:23 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/4/2021 1:37 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
>>
>>>>>>>> You are wrong because
>>>>>>>>       Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> where Linz requires that
>>>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>      if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>>
>>>>>>> You have the above incorrectly in these two
>>>>>>> M refers to the Turing Machine described by wM.
>>>>>>>
>>>>>>> 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
>>>>>>>
>>>>>>> This is *NOT* the Ĥ that is executed.
>>>>>>
>>>>>> I have no idea what you are objecting to.  You can substitute Ĥ
>>>>>> and ⟨Ĥ⟩
>>>>>> into what Linz says, can't you?  I did it above and you don't like
>>>>>> it:
>>>>>> For M = Ĥ and wM = ⟨Ĥ⟩ Linz tells us that
>>>>>>
>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>      if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>
>>>>>> Your Ĥ is not like that.  Its behaviour is boring and obvious and
>>>>>> trivial.
>>>>>
>>>>> This is how it really is:
>>>>> if ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>>>>> and ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its final state.
>>>> No.  ⟨Ĥ⟩ is a string, not a TM.  The TM that string represents is Ĥ and
>>>> Ĥ applied to ⟨Ĥ⟩ does halt.
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>> The question is not: Does Ĥ halt on its input?
>>
>> Yes it is. 
>
> The question is:
> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
> The ansswer to this question is provably no!
>
> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn  THIS IS NOT A CONTRADICTION
>

It is provably that it does.

Remember ALL copies of a given Turing Machine behave the same, so since
one copy of H^(H^) is show to halt, ALL copies of it are also proved to
Halt.

If you claim that this copy of H^(H^) doesn't halt then either:

The representation isn't correct, and your proof is thus bogus.

H^ isn't actually a Turing Machine, which implies that H isn't a Turing
Machine and your proof is thus bogus.

or

H^ isn't actual;y a Turing Machine because H^ wasn't built correctly
according the the Linz etc formula, and thus your proof is bogus.

Or, you proof is just bogus and it never proved what you claim it is.

Which one is it?

In my opinion, it really is all of the above.

On 8/5/21 9:04 AM, olcott wrote:

> André  refuses an honest dialogue
>

Seems to be the same for you.

On 8/5/2021 5:15 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 8/4/2021 6:22 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 8/4/2021 2:22 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 8/4/2021 7:53 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 8/2/2021 8:45 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>>>>> As long as it is simply dismissed out-of-hand as a contradiction the
>>>>>>>>>> paradox remains unresolved.
>>>>>>>>>
>>>>>>>>> There is no contradiction or paradox. You Ĥ is just the wrong sort of
>>>>>>>>> TM. The proof you want to "refute" is talking about this sort of Ĥ:
>>>>>>>>>
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>>>
>>>>>>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>>>>>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>>>>>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>>>>>>> Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
>>>>>>>
>>>>>>> Maybe saying it a couple more times will help. After four times I can
>>>>>>> tell you that it's still wrong. Maybe about a dozen more?
>>>>>>>
>>>>>>> Whether what happens after Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correct or not is determined
>>>>>>> by Linz, not by you. And you are clear that
>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn.
>>>>>>
>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>> if M applied to wM does not halt
>>>>>>
>>>>>> As explained in complete detail below:
>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> Yes, please don't tell me the final state yet again. This is not been
>>>>> in dispute for some time.
>>>>>
>>>>>> because M applied to wM does not halt
>>>>>> where M is Machine_of(⟨Ĥ⟩) (1st param) above
>>>>>> and wM is ⟨Ĥ⟩ the second param above.
>>>>>>
>>>>>> Because wM is referring to ⟨Ĥ⟩ and M is referring to the underlying
>>>>>> machine of ⟨Ĥ⟩ the last line above is translated to: if
>>>>>> Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ does not halt
>>>>> That's convoluted. ⟨Ĥ⟩ is the encoding of Ĥ so to find out what Linz
>>>>> expects from Ĥ applied to ⟨Ĥ⟩ we just substitute M = Ĥ and wM = ⟨Ĥ⟩ into
>>>>> the above:
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>
>>>> It is not the first Ĥ that is being referred to it is *only* the
>>>> machine represented by the input ⟨Ĥ⟩ that is being referred to.
>>> The machine represented by ⟨Ĥ⟩ is Ĥ. There is only one Ĥ being
>>> discussed here -- yours.
>>>
>>>> That machine
>>>> never reaches its final state.
>>> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn, or so you've told us. Are you changing
>>> your story? I'd like an answer. It's not a hard question.
>>
>> Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn
>
> OK, so you are not changing your story. Linz tells us that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.

The input to H will be the description (encoded in some form) of M, say
WM, as well as the input w. The requirement is then that, given any (WM,
w), the Turing machine H will halt with either a yes or no answer.
http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

The M that is being referred below is the machine specified by WM
H.q0 WM W ⊢* H.qn
if M applied to wM does not halt

The M that is being referred below is the machine specified by wM AKA ⟨Ĥ⟩
Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt

When Ĥ.qx is a simulating halt decider the question becomes:
Does the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ reach its final state.

> so your Ĥ is not interesting. Such TMs are ten-a-penny and none of them
> have anything to say about the proof in Linz. You should apologise for
> hooking people in with a lie 30 months ago. You don't have anything
> that anyone would consider to be impossible. What is impossible, and
> you once false claimed to have, is a TM that behaves as Linz says, if
> only for this one case.
>
>> because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.
>
> The reason is not relevant. That's good because the above is nonsense,
> iteral nonsense. What is the input "to"? It can't be Ĥ.qx because
> that's a state and states don't have input. It can't be Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
> because that's a TM configuration (state+tape contents) and these don't
> have input either.
>
> But it does not matter what this jumble of words means because Ĥ.q0 ⟨Ĥ⟩
> transitions to Ĥ.qn and it shouldn't.
>
>>>> Ĥ.q0 ⟨Ĥ[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
>>>> if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.
>>>>
>>>> That you persistently ignore this distinction really seems to be a
>>>> diverge from an honest dialogue.
>>>
>>> If the various strings you've chosen to number (badly) are not identical
>>> then your "hat" construction is wrong. Linz's "hat" version makes an
>>> exact copy.
>>
>> When Bill says that his identical twin brother is not going to go to
>> the store, and then Bill goes to the store this does not make him a
>> liar.
>
> Evasion. Is ⟨Ĥ[i]⟩ = ⟨Ĥ⟩ for all i or not? You need to start answering
> simple technical questions like this. If "yes", the notation is there
> only to help your exposition, if "no" your Ĥ is not behaving at it
> should.
>
> (Twins, unlike strings, are not mathematical objects, but if were to
> model twins mathematically we'd have to used non-equal objects since
> they can't be substituted one for the other.)
>
>>> You are free to write ⟨Ĥ[99]⟩ if you like, but I am also free to change
>>> that back to ⟨Ĥ⟩ because they are identical strings. Anything true of
>>
>> Ĥ is not a string it is a Turing machine.
>> Turing machines are not identical to strings.
>> Ĥ halts the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its final state
>> whether or not the halt decider at Ĥ.qx stop simulating it.
>
> There some words missing there I think. But it's irrelevant waffle.
> The undisputed fact is that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn but Linz's Ĥ
> should not behave like this. Everything that happens on the way to qn
> is irrelevant.
>
>> If you want an actual honest dialogue we must have closure on some of
>> the points. You must say what things you agree with and not merely
>> ignore those things that you agree with.
>
> Likewise. But your task is simpler than mine. We both agree that
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn. All you have to agree with is that this is not the
> behaviour specified by Linz: Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn if (and only if) Ĥ applied
> to ⟨Ĥ⟩ does not halt.
>
> On the other hand, I have to put pick waffle about input to something
> that has no input. Strings never halting or not reaching some state
> when string are just strings. All of which concerns the details of how
> your Ĥ is wrong.
>
>> Do you understand that the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its
>> final state whether or not the halt decider at Ĥ.qx stop simulating
>> it?
>
> No. I can't parse the sentence. To what does "its" refer? There is
> only one final state in Ĥ and it's Ĥ's final state!
>
> There's a deep problem here. You've never written a UTM (I'm not using
> "poetic license" here, I mean a UTM) so you don't know how a UTM
> simulation proceeds. The string is not decoded into a TM with its own
> states. It can't even use the tape in the usual way because the tape is
> now used for more than one purpose. Students get this (eventually)
> because they read the textbook and do the exercises.
>
> But the part you keep ignoring is why should anyone care, since Ĥ, and
> the H it's build from, do not behave as Linz describes?
>
>> If you disagree then to prove that you are not simply being
>> disagreeable you must proceed with a dialogue on this single point
>> until we reach mutual agreement. All other points will not be
>> discussed until we reach mutual agreement on this point.
>
> Nonsense. You've stated and agreed to the fact that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn.
> The details of that wrong journey are just something you want me to get
> bogged down in. You could simply agree with the plain fact that anyone
> could write an Ĥ that behaves like yours and none of them would say
> anything about the proof. You'd be free to go a walk some dogs.
>

olcott <NoOne@NoWhere.com> writes:

> The M that is being referred below is the machine specified by wM AKA ⟨Ĥ⟩

Yes, so M is Ĥ (what you wrote, rather redundantly, as Machine_Of(⟨Ĥ⟩).

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

So

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt

Linz even writes this out for you with the correct substitutions.

> When Ĥ.qx is a simulating halt decider the question becomes:
> Does the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ reach its final state.

Your implementation choices for H (and thus Ĥ) do not change the
specification. You Ĥ does not meet it.

We can go round and round this loop forever, but you can't show that
your Ĥ does what it should to be interesting.

--
Ben.

olcott <NoOne@NoWhere.com> writes:

> the Turing machine halting problem. Simply stated, the problem is:
> given the description of a Turing machine M and an input w, does M,
> when started in the initial configuration q0w, perform a computation
> that eventually halts? Using an abbreviated way of talking about the
> problem, we ask whether M applied to w, or simply (M, w), halts or
> does not halt.
> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>
> 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 Turing machine of the inputs not the Turing machine
> that is being executed.

M refers to the TM encoded as wM. The case in point -- the one you
claimed to have something important to say about -- has wM = ⟨Ĥ⟩ and
(consequently) M = Ĥ.

The key case is exactly the one where M refers to the TM being executed
because the input, wM, is ⟨Ĥ⟩. Linz writes it out for you (though I'll
switch to your notation):

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn if Ĥ applied to ⟨Ĥ⟩ does not halt.

For obvious reasons (if not, ask), you do not have such an Ĥ.

> if Machine_Of(wM) applied to wM does not halt

Yes, Machine_Of(wM) is M. This is saying what Linz is saying though it
a more wordy way.

> So when Bill says that his identical twin brother is never going to
> the store and then Bill goes to the store THIS IS NOT A CONTRADICTION.

There is no contradiction. I've said this many times now. There is no
contradiction (or paradox) with your Ĥ.

Mind you, I don't see what your analogy is trying to tell me so I
suspect you are wrong about something here. What are the two things
that look identical but do different things here? Bill and his brother
are... what... Ĥ and Ĥ or ⟨Ĥ⟩ and ⟨Ĥ⟩?

> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn THIS IS NOT A CONTRADICTION.

<Sigh> How often do I have to say something for you take note of it?
There is no contradiction or paradox with your Ĥ. Your Ĥ is simply not
doing what Linz says it should be doing to be interesting. It's doing
the first part of the spec: Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn but it's not meeting the
second part because, obviously, Ĥ applied to ⟨Ĥ⟩ halts.

If, 30 months ago, you'd said "I have two TMs H and Ĥ (as per Linz's
construction) with this behaviour: Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn" no
one would have cared. I don't know if you genuinely thought you had
something interesting, but it 100% clear now that you never did.

The false claim you made was that your H/Ĥ pair were not just
constructed as in Linz, but met Linz's specifications, at least for the
one key case.

> If you don't want an honest dialogue then please say so.

I can't imagine what you think I am being less than honest about. You,
on the other hand have said things like:

"Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz
specs does not exist. I now have a fully encoded pair of Turing
Machines H / Ĥ proving them wrong."

Now you may have been deluded or just mistaken when you said this, but
you must see, now, that you don't have anything meeting Linz's
specification. It's dishonest to keep saying (or implying) that you do.

--
Ben.

olcott <NoOne@NoWhere.com> writes:

> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>> The question is not: Does Ĥ halt on its input?
>> Yes it is.
>
> The question is:
> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
> The ansswer to this question is provably no!

The question is: does Ĥ applied to ⟨Ĥ⟩ halt. It does:

> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn THIS IS NOT A CONTRADICTION

Indeed. There is no contradiction. Just an Ĥ that does not meet Linz
spec.

--
Ben.

On 8/5/2021 9:35 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> The M that is being referred below is the machine specified by wM AKA ⟨Ĥ⟩
>
> Yes, so M is Ĥ (what you wrote, rather redundantly, as Machine_Of(⟨Ĥ⟩).
>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>> if M applied to wM does not halt
>
> So
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if Ĥ applied to ⟨Ĥ⟩ does not halt
>
> Linz even writes this out for you with the correct substitutions.
>
>> When Ĥ.qx is a simulating halt decider the question becomes:
>> Does the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ reach its final state.
>
> Your implementation choices for H (and thus Ĥ) do not change the
> specification. You Ĥ does not meet it.
>
> We can go round and round this loop forever, but you can't show that
> your Ĥ does what it should to be interesting.
>

That you ignore rather than point out errors in my reasoning really
seems to prove that you don't want an honest dialogue.

If you really do want an honest dialogue then we must go through the
points on page 6 until we have mutual agreement.
Page 6 conclusively proves that H(P,P)==0 is correct.

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