Torkel Franzen on truth
in [Logic]
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From: george on 2 Jan 2008 10:26 On Jan 1, 9:14 am, Aatu Koskensilta <aatu.koskensi...(a)xortec.fi> wrote: > Bah. We could with equal justification, and equally arbitrarily, claim that > "the completeness theorem is fundamentally a proof that first-order syntactic > deduction SIMPLY DOESN'T EXIST". Since the completeness theorem is, itself, LIKE ALL theorems, an exercise IN AND USING syntactic deduction that is NOT going to have "equal" justification. And none of it is arbitrary to begin with. You are just fulminating. Incompetently.
From: george on 2 Jan 2008 10:32 On Jan 1, 9:14 am, Aatu Koskensilta <aatu.koskensi...(a)xortec.fi> wrote: > Bah. We could with equal justification, and equally arbitrarily, claim that > "the completeness theorem is fundamentally a proof that first-order syntactic > deduction SIMPLY DOESN'T EXIST". The theorem wasn't attempted in that direction. It was NOT like we had a PRIOR definition of logical consequence (over infinite domains, with finitary machinery) and were proving that we could re-cast it another way (model theoretically). The original Tarskian definition is model-theoretic. The question is whether any finitary syntactic system can "completely" emulate it. All of your allegation of absurdity here is founded on the superficial observation that proving an equivalence between two things "cannot" be proving that either of them doesn't exist. Nobody is impressed. You are basically flaunting an unwillingness (NObody in the target audience, especially not you, is actually stupid enough to flaunt inability) to apply the relevant metaphor.
From: george on 2 Jan 2008 10:36 On Dec 21 2007, 1:37 am, herbzet <herb...(a)gmail.com> wrote: > If you consider the sign for equality a logical symbol, its meaning > will be conferred by the axioms concerning it. If it is a *logical* symbol then it will necessarily be *prior* to any axioms. In FOL with equality, substitutivity in particular *should* be classified as a rule of inference AS OPPOSED to an axiom-schema. The Logic is characterized by its ROI's and in FOL *with* equality, the equality is part of the logic.
From: george on 2 Jan 2008 10:45 On Dec 21 2007, 4:03 am, Peter_Smith <ps...(a)cam.ac.uk> wrote: > If I tell you that "mae glo yn du" is true in Welsh [heck, hope I've > remembered that right], you don't thereby get to know what it means. > Telling you the same about some other Welsh sentences won't help > either. There are logical symbols as well. They provide a sort of skeleton for the structure. In particular, IF Welsh has a word for "not", then it arguably IS possible, JUST from surveying the true sentences, to discover it.
From: george on 2 Jan 2008 11:02
On Dec 7 2007, 5:59 pm, Peter_Smith <ps...(a)cam.ac.uk> wrote: > But again, what *I* said was that the canonical > Gödel sentence *of a sound theory* will be > true on the standard interpretation. This is begging all sorts of questions and mixing all sorts of notions. There is a PRIOR question that you are continuing to lamely dodge and that Isaacson didn't even realize he was confronting: WHAT DOES IT *mean*, in GENERAL, for a "theory" to BE "sound"?? In the case of arithmetic, here, THERE ALREADY IS a standard interpretation. A standard MODEL. ANY theory over the same language as those axioms is sound BY DEFINITION if AND ONLY IF it AGREES with the standard interpretation! THAT IS THE DEFINITION of "sound"! > That follows immediately, OF COURSE it follows immediately that any sentence in any sound theory is true in the standard model, SINCE "sound" is DEFINED as "matching the standard model"! THAT IS *NOT* the point! > 1) For any theory S, the canonical Gödel sentence > G_S is true on the > standard interpretation if and only if S doesn't prove G_S. [By > construction] You have to go to ZF to say ANYthing whatsoEVER about "the standard interpretation". Good grief. What is constructible will occur IN THE NONSTANDARD INTERPRETATIONS AS WELL. > and > > 2) For any sound theory S, > S doesn't prove G_S [By Gödel's semantic > argument for incompleteness] Why do you say "sound" here? For ANY theory rich enough to formulate Con(S) or G_S at all, S doesn't prove G_S. |