'When Are Relations Neither True Nor False? [Logic]

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From: Jesse F. Hughes on 26 Feb 2010 06:51

Newberry <newberryxy(a)gmail.com> writes:

> On Feb 25, 11:33 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
>> Let's ask another question, one which you have never answered as far
>> as I recall. I do not know whether Goldbach's conjecture is true or
>> false, but I do know that there are no counterexamples to GC that are
>> less than, say, 27. In fact, this is easy to check. Thus, it seems
>> to me perfectly reasonable to say that:
>>
>> (Ax)( if x is a counterexample to GC, then x >= 27 )
>>
>> That's a true statement, as far as I'm concerned.
>
> True statement is this
>
> ~T[(Ex)((x is a counterexample to GC) & (x < 27))

That stands for, I suppose, "It is not true that ..."?

And, according to what you wrote below, it's a second-order statement?

This prompts lots of questions, of course, but I'll start with a
simple one. The reason that you introduce the T predicate is that you
think the statement

~(Ex)((x is a counterexample to GC) & (x < 27)) (*)

might be meaningless, right? And yet, we *use* that statement in
simple derivations. Is 8 a counterexample to GC? No, because we know
that there are no counterexamples below 27. That is a perfectly
sensible inference, using a statement that you think might be
meaningless.

Does that not bother you?

By the way, refresh my memory. If (*) is meaningless, what about
~(*)? Do you call that meaningless or false?

In any case, the proof of (*) is relatively simple and uses no
second-order stuff. The sentence (*) is plainly meaningful and I
daresay that *you know what it means*. It means that there are no
counterexamples to GC less than 27. But you have to pretend that
you're not sure whether this statement is true or meaningless and
hence have to pretend that a perfectly simple proof in FOL is a proof
of a second-order statement about a possibly meaningless statement.

Does that sound about right?

>> How about you? You don't know whether it's true or meaningless unless
>> you know whether there are counterexamples to GC.
>
> True.
>
>> Thus, you can't say
>> whether my proof is a real proof or not, until you determine whether
>> GC is true.
>
> Yes, I can. Please refer to my second order statement above.
>
>> If GC is true, then my proof must not be a proof (since
>> its conclusion is meaningless). If GC is false, then my proof is a
>> proof after all.
>>
>> Is this sensible to you? Is my "theorem" true or meaningless or are
>> you unable to decide which?

--
Jesse F. Hughes
"I get to make things move just by saying a few things. When I post
now the math world has to tremble, even if it does so quietly, hoping
that no one else notices." -- James S. Harris has the power.

From: Daryl McCullough on 26 Feb 2010 07:10

Newberry says...
>
>On Feb 25, 8:18=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
>wrote:
>> Newberry says...

>> >This is what Torkel Franzen said. Look, Frege also believed that the
>> >axioms in his system were manifestly true.
>>
>> Well, he was mistaken. The belief that a system is consistent
>> can be mistaken.
>
>Which of his axioms was not manifestly true?

Unrestricted comprehension, which says if Phi(x) is a formula
with one free variable, then there is a set y such that forall
x,

x is an element of y
<->
Phi(x)

This is clearly not true in the case Phi(x) is the formula
"x is an element of x".

--
Daryl McCullough
Ithaca, NY

From: Daryl McCullough on 26 Feb 2010 07:12

Newberry says...
>
>On Feb 25, 7:44=A0am, Frederick Williams <frederick.willia...(a)tesco.net>
>wrote:

>> A theory is sound if all its theorem are true in all models (of the
>> appropriate type).
>
>It seems to me that it follows that what a sound theory can prove
>depends on what you consider true, does it not?

That's a weird way of putting it. I would say, rather, that
"It follows that whether a theory is sound or not depends on
what you consider true".

--
Daryl McCullough
Ithaca, NY

From: Newberry on 26 Feb 2010 09:55

On Feb 26, 3:51 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> Newberry <newberr...(a)gmail.com> writes:
> > On Feb 25, 11:33 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> >> Let's ask another question, one which you have never answered as far
> >> as I recall. I do not know whether Goldbach's conjecture is true or
> >> false, but I do know that there are no counterexamples to GC that are
> >> less than, say, 27. In fact, this is easy to check. Thus, it seems
> >> to me perfectly reasonable to say that:
>
> >> (Ax)( if x is a counterexample to GC, then x >= 27 )
>
> >> That's a true statement, as far as I'm concerned.
>
> > True statement is this
>
> > ~T[(Ex)((x is a counterexample to GC) & (x < 27))
>
> That stands for, I suppose, "It is not true that ..."?

Obviously.

> And, according to what you wrote below, it's a second-order statement?
>
> This prompts lots of questions, of course, but I'll start with a
> simple one. The reason that you introduce the T predicate is that you
> think the statement
>
> ~(Ex)((x is a counterexample to GC) & (x < 27)) (*)
>
> might be meaningless, right?

Right.

And yet, we *use* that statement in
> simple derivations.

We use it as a shorthand. If we want to be rigourous we have to use
~T(). Sometimes we just tend to conflate the two. Anyway, my point is
that we can express what we need to express only with finer
granularity.

> Is 8 a counterexample to GC? No, because we know
> that there are no counterexamples below 27. That is a perfectly
> sensible inference, using a statement that you think might be
> meaningless.
>
> Does that not bother you?

No. We just got used to conflating two different things.

>
> By the way, refresh my memory. If (*) is meaningless, what about
> ~(*)? Do you call that meaningless or false?

For the Almighty's sake, a negation of a meaningles sentence mus be
meaningless.

> In any case, the proof of (*) is relatively simple and uses no
> second-order stuff. The sentence (*) is plainly meaningful and I
> daresay that *you know what it means*.

I know exactly what you want to say: ~T[(Ex)((x is a counterexample to
GC) & (x < 27)). You are just sloppy in expressing it.

> It means that there are no
> counterexamples to GC less than 27. But you have to pretend that
> you're not sure whether this statement is true or meaningless and
> hence have to pretend that a perfectly simple proof in FOL is a proof
> of a second-order statement about a possibly meaningless statement.
>
> Does that sound about right?

No. I am not pretending.

>
>
>
>
>
> >> How about you? You don't know whether it's true or meaningless unless
> >> you know whether there are counterexamples to GC.
>
> > True.
>
> >> Thus, you can't say
> >> whether my proof is a real proof or not, until you determine whether
> >> GC is true.
>
> > Yes, I can. Please refer to my second order statement above.
>
> >> If GC is true, then my proof must not be a proof (since
> >> its conclusion is meaningless). If GC is false, then my proof is a
> >> proof after all.
>
> >> Is this sensible to you? Is my "theorem" true or meaningless or are
> >> you unable to decide which?
>
> --
> Jesse F. Hughes
> "I get to make things move just by saying a few things. When I post
> now the math world has to tremble, even if it does so quietly, hoping
> that no one else notices." -- James S. Harris has the power.- Hide quoted text -
>
> - Show quoted text -

From: Newberry on 26 Feb 2010 09:56

On Feb 26, 4:10 am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
wrote:
> Newberry says...
>
>
>
> >On Feb 25, 8:18=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
> >wrote:
> >> Newberry says...
> >> >This is what Torkel Franzen said. Look, Frege also believed that the
> >> >axioms in his system were manifestly true.
>
> >> Well, he was mistaken. The belief that a system is consistent
> >> can be mistaken.
>
> >Which of his axioms was not manifestly true?
>
> Unrestricted comprehension, which says if Phi(x) is a formula
> with one free variable, then there is a set y such that forall
> x,
>
> x is an element of y
> <->
> Phi(x)
>
> This is clearly not true in the case Phi(x) is the formula
> "x is an element of x".

Looks like you discovered that it was not manifestly true only AFTER
you derived a contradiction.