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From: Daryl McCullough on 13 Aug 2010 07:17
Newberry says...
>
>On Aug 12, 10:15=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
>wrote:

>> Well, in the case of arithmetic, presumably everything that is true
>> is necessarily true, right? Everything that is false is necessarily
>> false.
>
>Right, that is why
>
>>~(Qm & (Ex)Pxm)
>
>is ~(T v F) if (Ex)Pxm is false. I recommend reading section 2.2.
>Truth-relevance is a very simple concept.

Which appears to be completely useless for mathematics.

--
Daryl McCullough
Ithaca, NY

From: Newberry on 13 Aug 2010 09:02
On Aug 13, 4:03 am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
wrote:
> Newberry says...
>
>
>
>
>
>
>
> >On Aug 12, 8:35=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
> >wrote:
> >> That's just bizarre. With the interpretation that Qm holds
> >> only if m is the Godel number of the Godel sentence G, then
> >> (Pxm & Qm) says
>
> >> "x is a code for a proof of the formula whose code is m
> >> and m is the code for G"
>
> >> which is just an indirect way of saying
> >> "x is code for a proof of G".
>
> >> So ~(Ex) (Pxm & Qm)
>
> >> is an indirect way of saying "There is no proof of G".
>
> >> Calling it vacuous is just bizarre. Why in the world would
> >> you want to do that?
>
> >Because ~(Ex)Pxm. The sentence above is equivalent to
>
> >(x)(Pxm -> ~Qm)
>
> >It is clearly "vacuously true" (in classcal logic), is it not?
>
> It seems to me that all true formulas of arithmetic are
> vacuously true. It's not an interesting concept.

The formulas of the form

(x)(Ax -> Bx)

are vacuous if ~(Ex)Ax. I do not think this is the case for all
formulas. For example

2 + 2 = 4 or (x)(x + 1 > x)

are not vacuous.

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

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