Torkel Franzen on truth
in [Logic]
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From: Daryl McCullough on 20 Nov 2007 09:38 Newberry says... >On Nov 19, 11:38 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> How are you defining T(F)? Tarski showed that no consistent >> language can have a truth predicate for that very language. > >I am not defining it. You defined it in your previous posts. No, I didn't. I used the symbol T to mean a particular theory. So by T(F) do you mean "T proves F"? >I did not say that it was in the language of PA. I said that >T(F) --> F >is just as compelling as any other axiom. Axiom for *what*? Yes, if you start with a theory T, you can use the soundness schema for T as an axiom for a new theory T1. You can't use it as an axiom schema for T, because that's ill-defined. You have to know what T is before you can add T(F) -> F as an axiom. >What I mean is that by puting T(F) on another level >you did avoid a contradiction, nevertheless intuitively >T(F) --> F. ("F" means any wff not "false."} Yes, if T is sound, then T(F) -> F (that's what "sound" means). There is no problem with adding such an axiom. -- Daryl McCullough Ithaca, NY
From: Newberry on 20 Nov 2007 10:47 On Nov 20, 6:38 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Newberry says... > > >On Nov 19, 11:38 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >wrote: > >> How are you defining T(F)? Tarski showed that no consistent > >> language can have a truth predicate for that very language. > > >I am not defining it. You defined it in your previous posts. > > No, I didn't. I used the symbol T to mean a particular theory. > So by T(F) do you mean "T proves F"? "T(F)" means "F is true" > > >I did not say that it was in the language of PA. I said that > >T(F) --> F > >is just as compelling as any other axiom. > > Axiom for *what*? If a sentence F is true then F. Yes, if you start with a theory T, > you can use the soundness schema for T as an axiom for > a new theory T1. You can't use it as an axiom schema for > T, because that's ill-defined. You have to know what T > is before you can add T(F) -> F as an axiom. > > >What I mean is that by puting T(F) on another level > >you did avoid a contradiction, nevertheless intuitively > >T(F) --> F. ("F" means any wff not "false."} > > Yes, if T is sound, then T(F) -> F (that's what "sound" > means). There is no problem with adding such an axiom. > > -- > Daryl McCullough > Ithaca, NY
From: Daryl McCullough on 20 Nov 2007 11:58 Newberry says... >On Nov 20, 6:38 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >"T(F)" means "F is true" Well, there is no such formula that means "F is true" for *arbitrary* F. You can restrict yourself to a specific language L and introduce a predicate T(F) with the interpration that T(F) holds if F is a true formula of language L, but as proved by Tarski, the predicate T cannot be in the language L. >> >I did not say that it was in the language of PA. I said that >> >T(F) --> F >> >is just as compelling as any other axiom. >> >> Axiom for *what*? > >If a sentence F is true then F. I meant: axiom for what theory? -- Daryl McCullough Ithaca, NY
From: LordBeotian on 20 Nov 2007 14:13 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> ha scritto >>"T(F)" means "F is true" > > Well, there is no such formula that means "F is true" > for *arbitrary* F. You can restrict yourself to a specific > language L and introduce a predicate T(F) with the interpration > that T(F) holds if F is a true formula of language L, but > as proved by Tarski, the predicate T cannot be in the language > L. Are you saying that - adding to the language L of PA a unary relation symbol "T" to obtain the language L' - adding to PA the axiom schemata "T(#F)->F" for any wff F of L' we get an inconsistent theory?
From: Daryl McCullough on 20 Nov 2007 14:58
LordBeotian says... >"Daryl McCullough" <stevendaryl3016(a)yahoo.com> ha scritto > >>>"T(F)" means "F is true" >> >> Well, there is no such formula that means "F is true" >> for *arbitrary* F. You can restrict yourself to a specific >> language L and introduce a predicate T(F) with the interpration >> that T(F) holds if F is a true formula of language L, but >> as proved by Tarski, the predicate T cannot be in the language >> L. > >Are you saying that >- adding to the language L of PA a unary relation symbol "T" to obtain the >language L' >- adding to PA the axiom schemata "T(#F)->F" for any wff F of L' >we get an inconsistent theory? Actually, that schema by itself is consistent. That schema doesn't say that T is the truth predicate. It only says that T is sound. You need T(#F) <-> F in order for T to be the truth predicate. Adding that schema leads to an inconsistency. -- Daryl McCullough Ithaca, NY |