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

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From: Newberry on 24 Mar 2010 00:48

On Mar 23, 6:10 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote:
> stevendaryl3...(a)yahoo.com (Daryl McCullough) writes:
> > Well, Jesse's statement can be rewritten in the form
>
> > (forall a,b)( a^2/b^2 = 2 --> gcd(a,b) > 1 )
>
> But does classical equivalence preserve meaninglessness?

No. For example uniform substitution does not preserve meaningfulness
(or T-relevance in Diaz's terminology.)
>
> --
> Aatu Koskensilta (aatu.koskensi...(a)uta.fi)
>
> "Wovon man nicht sprechan kann, darüber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus

From: Jesse F. Hughes on 24 Mar 2010 06:24

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

> On Mar 23, 5:58 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
>> Aatu Koskensilta <aatu.koskensi...(a)uta.fi> writes:
>> > "Jesse F. Hughes" <je...(a)phiwumbda.org> writes:
>>
>> >> Thus. the sentence
>>
>> >> ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 )
>>
>> >> is meaningless, right?
>>
>> > Newberry said that (x)(Px --> Qx) is meaningless if ~(Ex)Px is
>> > necessarily true. How do you get from this the meaninglessness of
>>
>> > ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 )
>>
>> > which is not of the form (x)(Px --> Qx)?
>>
>> He also says that
>>
>> ~(Ex)(Px & Qx)
>>
>> is meaningless in exactly the same situations that
>>
>> (Ax)(Px -> Qx)
>>
>> is meaningless. He wants the two formulas to remain equivalent.
>>
>> (Surely, you're not taking issue with the fact that I've used two
>> existential statements rather than one?)
>>
>> As usual, Newberry can correct me if I'm mistaken on his claims.
>
> Yes, this is what I am proposing. Not sure what two existential
> statements you are referring though.

I meant two existential quantifiers (There exists a and b such that
....) rather than statements. But no matter. As we all know, that's
essentially the same as a single quantifier.

--
"If .999... = 1 then (.999...)/1 should equal 1
let's see
(.999...)/1 = .999...
[Therefore] .999... still=/= 1" -- An astonishing proof by "S. Enterprize"

From: Jesse F. Hughes on 24 Mar 2010 06:32

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

> Plus
>
> (x)((x = x + 1) -> (x = x + 2))
>
> does not look particularly meaningful to me.

I don't believe you. You know what it means. It's perfectly clear
what it means. It means that whenever x = x + 1, then x = x + 2.[1]

Just as it's perfectly clear what "There are no counterexamples to
Goldbach's conjecture that are less than 12." You and I both know
what this means and we won't suddenly be struck stupid if it turns out
that Goldbach's conjecture is true.

Your claims about the meaninglessness of simple arithmetic statements
are among the least plausible.

Footnotes:
[1] In fact, this statement seems obviously true! Suppose
x = x + 1. Then we may substitute x + 1 for x in the right hand side
of the equation x = x + 1, thus:

x = x + 1
= (x + 1) + 1
= x + 2.

I see nothing the least bit fishy about this reasoning.

--
"It's one of the easiest tickets to true fame--not this silly stuff
where people cheer you for a few years and then forget about you--but
the kind of fame where school kids have to read your biography and do
reports on you." -- Another reason to support James S. Harris.

From: Daryl McCullough on 24 Mar 2010 06:58

Newberry says...
>
>On Mar 23, 5:42=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
>wrote:

>> The problem with this talk about statements being "meaningless" is
>> that often the only way that you *know* that ~(Ex)Px is necessarly true
>> is by proving a pair of statements:
>>
>> 1. (Ax) Px -> Qx
>>
>> 2. (Ax) Px -> ~Qx
>>
>> together, these imply
>>
>> 3. ~(Ex) Px
>>
>> which you claim implies that 1&2 are meaningless.
>
>Maybe it is like a proof by contradiction. You assume (Ex)Px and
>derive ~(Ex)Px.
>
>How do you know that there are no alternative proofs?

I don't know that there are no alternative proofs, but I
know that the proof using 1-3 works perfectly fine. I can't
see any *advantage* to calling some statements meaningless,
especially when you can derive true consequences from them.

>> I think it makes more sense to say that every statement
>> of arithmetic is meaningful.
>
>No, system with gaps has better properties than a system without gaps.

Better in what sense?

>Plus
>
> (x)((x = x + 1) -> (x = x + 2))
>
>does not look particularly meaningful to me.

The classical meaning of an implication "A implies B"
is that it is *not* the case that A is true and B is false.
So for A implies B to be true, there are three possibilities:

1. A is true and B is true.
2. A is false and B is true.
3. A is false and B is false.

In the case x=x+1 -> x=x+2, both are false, so we are
in case 3. There is nothing meaningless about this. It's
not a particularly *interesting* observation. But who
cares?

--
Daryl McCullough
Ithaca, NY

From: Newberry on 24 Mar 2010 12:11

On Mar 24, 3:58 am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
wrote:
> Newberry says...
>
>
>
>
>
>
>
> >On Mar 23, 5:42=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
> >wrote:
> >> The problem with this talk about statements being "meaningless" is
> >> that often the only way that you *know* that ~(Ex)Px is necessarly true
> >> is by proving a pair of statements:
>
> >> 1. (Ax) Px -> Qx
>
> >> 2. (Ax) Px -> ~Qx
>
> >> together, these imply
>
> >> 3. ~(Ex) Px
>
> >> which you claim implies that 1&2 are meaningless.
>
> >Maybe it is like a proof by contradiction. You assume (Ex)Px and
> >derive ~(Ex)Px.
>
> >How do you know that there are no alternative proofs?
>
> I don't know that there are no alternative proofs, but I
> know that the proof using 1-3 works perfectly fine. I can't
> see any *advantage* to calling some statements meaningless,
> especially when you can derive true consequences from them.
>
> >> I think it makes more sense to say that every statement
> >> of arithmetic is meaningful.
>
> >No, system with gaps has better properties than a system without gaps.
>
> Better in what sense?

It can solve the Liar paradox. Tarki's theorem does not apply to it.

>
> >Plus
>
> > (x)((x = x + 1) -> (x = x + 2))
>
> >does not look particularly meaningful to me.
>
> The classical meaning of an implication

The classical "meaning of an implication is known to be paradixical.

> "A implies B"
> is that it is *not* the case that A is true and B is false.
> So for A implies B to be true, there are three possibilities:
>
> 1. A is true and B is true.
> 2. A is false and B is true.
> 3. A is false and B is false.
>
> In the case x=x+1 -> x=x+2, both are false, so we are
> in case 3. There is nothing meaningless about this. It's
> not a particularly *interesting* observation. But who
> cares?

Right on! No harm will be done to arithmetic if such sentences are not
provable.

>
> --
> Daryl McCullough
> Ithaca, NY- Hide quoted text -
>
> - Show quoted text -