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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 - |