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

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From: Aatu Koskensilta on 1 Apr 2010 08:59

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

> Let
>
> ~(Ex)(Ey)(Pxy & Qy) (1)
>
> be Goedel's sentence. [Pxy: x is a proof of y, Qy is such that only
> one y=m satisfies it & m = #(1)]. You are absolutely convinced that PA
> is consistent i.e. (1) is true. That is, the search for x and y will
> never terminate. How do you know that the search for x and y will
> never terminate?

How do I know that Peano arithmetic is consistent? I know it the way I
know any mathematical theorem I have personally proved. Perhaps you'd
now be willing to say whether you agree that for any formal theory T
extending Robinson arithmetic, either directly or through an
interpretation, there are infinitely many true statements (of the form
"the Diophantine equation D(x1, ..., xn) = 0 has no solutions") which
are unprovable in T if T is consistent?

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

From: Jesse F. Hughes on 1 Apr 2010 09:05

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

> On Mar 31, 5:57 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
>> But that is certainly not the argument that was offered in the other
>> thread. The statement
>>
>> (En)(Ea,b,c) P(a,b,c,n) -> (En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) )
>>
>> was proved independently of your assumption, and no one balked. Thus,
>> my question remains: was the argument I gave invalid?
>
> It is valid but it makes a silent assumption.

An assumption so silent that no one but you noticed it?

>> A second question comes to mind: what happened to your imagination
>> test? You've said that (because FLT is true) you cannot picture
>> (E a,b,c,n)( P(a,b,c,n) & Q(a,b,c) ). Yet, once you assume (contrary
>> to fact) that FLT is false, you *can* picture it?
>
> No.
>>
>> You can't picture a green round triangle, right? What if I say:
>> assume a round triangle exists. Can you picture a green round
>> triangle *then*?[1]
>
> No.
>
>> That is, does the act of making an assumption
>> change your capacities for imagining stuff?
>
> No.
>
>> If not, then your
>> argument above doesn't work.
>
> Why not. Assume there are such things as meaningless sentences. These
> sentences are well formed. Well formed sentences can go through
> syntactic transformations.

A similar syntactic transformation proves

~ (En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) ) (1)

This syntactic transformation does not apparently involve any
assumptions at all. It seems to me that you're coming close to
advocating an unsound theory: one which proves (1), despite the fact
that you've said that (1) is not true.

In any case, I really think you ought to think these things through.
Up until now, you seem to have assumed that statements like (1) are
useless if they're vacuously true. Now you've seen several proofs
involving statements like (1) -- proofs which entail that these
statements *are* vacuously true!

It seems to me that quick responses to this observation are unlikely
to work. Why not take the time and figure out what you think counts
as a proof and see where that takes you?

--
Jesse F. Hughes
.... one of the main causes of the fall of the Roman Empire was that,
lacking zero, they had no way to indicate successful termination of
their C programs. -- Robert Firth

From: Nam Nguyen on 1 Apr 2010 20:30

Aatu Koskensilta wrote:

>
> How do I know that Peano arithmetic is consistent? I know it the way I
> know any mathematical theorem I have personally proved.

So what you're saying is you just _intuit_ PA system be consistent,
no more no less. Of course anyone else could intuit the other way too!

From: Newberry on 2 Apr 2010 00:32

On Apr 1, 6:05 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> Newberry <newberr...(a)gmail.com> writes:
> > On Mar 31, 5:57 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> >> But that is certainly not the argument that was offered in the other
> >> thread. The statement
>
> >> (En)(Ea,b,c) P(a,b,c,n) -> (En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) )
>
> >> was proved independently of your assumption, and no one balked. Thus,
> >> my question remains: was the argument I gave invalid?
>
> > It is valid but it makes a silent assumption.
>
> An assumption so silent that no one but you noticed it?
>
>
>
>
>
> >> A second question comes to mind: what happened to your imagination
> >> test? You've said that (because FLT is true) you cannot picture
> >> (E a,b,c,n)( P(a,b,c,n) & Q(a,b,c) ). Yet, once you assume (contrary
> >> to fact) that FLT is false, you *can* picture it?
>
> > No.
>
> >> You can't picture a green round triangle, right? What if I say:
> >> assume a round triangle exists. Can you picture a green round
> >> triangle *then*?[1]
>
> > No.
>
> >> That is, does the act of making an assumption
> >> change your capacities for imagining stuff?
>
> > No.
>
> >> If not, then your
> >> argument above doesn't work.
>
> > Why not. Assume there are such things as meaningless sentences. These
> > sentences are well formed. Well formed sentences can go through
> > syntactic transformations.
>
> A similar syntactic transformation proves
>
> ~ (En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) ) (1)
>
> This syntactic transformation does not apparently involve any
> assumptions at all. It seems to me that you're coming close to
> advocating an unsound theory: one which proves (1), despite the fact
> that you've said that (1) is not true.
>
> In any case, I really think you ought to think these things through.
> Up until now, you seem to have assumed that statements like (1) are
> useless if they're vacuously true. Now you've seen several proofs
> involving statements like (1) -- proofs which entail that these
> statements *are* vacuously true!
>
> It seems to me that quick responses to this observation are unlikely
> to work. Why not take the time and figure out what you think counts
> as a proof and see where that takes you?

Glad we finally agree on something. I do not know if assuming
existence i.e. a presupposition of another formula, then drawing a
conclusion from the formula that contradicts the initial existential
assumption is a valid argument. But I am not convinced that it is not
invalid either. Of course these things need to be worked out. I
actually started to work on a derivation system for predicate
calculus, basically trying to mimic the traditional syllogism as
Strawson has shown that the logic of prsuppositions and the
traditional syllogism are compatible or identical. I was thinking of
appending it to my t-relevant logic exposition paper. But it is not
anywhere near publishable stage. There are still a few issues to be
worked out. But we will get there in due course. It may not even be
that difficult.

I do not know how it will turn out. I forgot who proved that the
square root of 2 was irrational and what his proof looked like. Maybe
your version is something concocted by the modern mathematicians who
take classical logic for granted. Maybe it will turn invalid, maybe
valid with some modifications or added assumptions. Mind you the
Greeks did not have the concept that the vacuous sentences were true.
The traditional syllogism presupposes that the subject class is non-
empty.

>
> --
> Jesse F. Hughes
> ... one of the main causes of the fall of the Roman Empire was that,
> lacking zero, they had no way to indicate successful termination of
> their C programs. -- Robert Firth- Hide quoted text -
>
> - Show quoted text -

From: Newberry on 2 Apr 2010 00:33

On Apr 1, 5:59 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote:
> Newberry <newberr...(a)gmail.com> writes:
> > Let
>
> > ~(Ex)(Ey)(Pxy & Qy) (1)
>
> > be Goedel's sentence. [Pxy: x is a proof of y, Qy is such that only
> > one y=m satisfies it & m = #(1)]. You are absolutely convinced that PA
> > is consistent i.e. (1) is true. That is, the search for x and y will
> > never terminate. How do you know that the search for x and y will
> > never terminate?
>
> How do I know that Peano arithmetic is consistent? I know it the way I
> know any mathematical theorem I have personally proved.

You proved PA consistent?

> Perhaps you'd
> now be willing to say whether you agree that for any formal theory T
> extending Robinson arithmetic, either directly or through an
> interpretation, there are infinitely many true statements (of the form
> "the Diophantine equation D(x1, ..., xn) = 0 has no solutions") which
> are unprovable in T if T is consistent?
>
> --
> Aatu Koskensilta (aatu.koskensi...(a)uta.fi)
>
> "Wovon man nicht sprechan kann, dar ber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus