Simple yet Profound Metatheorem
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From: H. J. Sander Bruggink on 20 Dec 2005 09:14 Charlie-Boo wrote: > H. J. Sander Bruggink wrote: >>But you did talk about "truth tables" and "resolution", both >>of which are proof methods for classical propositional logic, >>and not intuitionistic propositional logic. > > Please give a proof of a propositional calculus proposition that cannot > be done using case analysis (examination of the truth tables.) Man, just admit that you don't know what intuitionistic logic is. You don't mention the word "intuitionistic" in your question anymore, making it ambiguous in the present context. If I interpret your question as: "Please give an intuitionistic proof of an (intuitionistically valid) propositional calculus proposition that cannot be done using case analysis." then the answer would be: *any proof*, because "case analysis" doesn't work for intuitionistic logic. groente -- Sander
From: G. Frege on 20 Dec 2005 09:23 On 20 Dec 2005 05:25:27 -0800, "Charlie-Boo" <chvol(a)aol.com> wrote: >> >> But you did talk about "truth tables" and "resolution", both >> of which are proof methods for classical propositional logic, >> and not intuitionistic propositional logic. >> > Please give a proof of a propositional calculus proposition that > cannot be done using case analysis (examination of the truth tables). > Theorem (of PC): ~~P -> P Proof: (1) ~~P A (2) P 1 ~E (3) ~~P -> P 1,2 ->I Using truth tables you would determine that "~~P -> P" is a tautology, I guess. But still "~~P -> P" is _not_ a theorem of intuitionistic propositional logic. A. -- "I do tend to feel Hughes & Cresswell is a more authoritative source than you." (David C. Ullrich)
From: G. Frege on 20 Dec 2005 09:44 On 20 Dec 2005 05:35:47 -0800, "Charlie-Boo" <chvol(a)aol.com> wrote: * Thm. ((|- P) = (|- Q)) => |- (P = Q) (Charlie-Boo) >>>> >>>> This is obviously false. >>>> If neither P nor Q is provable (so that >>>> ((|- P) = (|- Q)) is true) it certainly >>>> does not follow that P==Q _is_ provable. >>>> >>> Let's suppose that P is "1>2" and Q is "2>3". Then P==Q is >>> (1>2)==(2>3) which IS provable. >>> >> Huh? > > I said, let's suppose that P is "1>2" and Q is "2>3". Then P==Q is > (1>2)==(2>3) which IS provable. > Man, it's n o t David C. Ullrich's problem that y o u are completely clueless concerning mathematical logic. Your "Theorem" is a _general_ statement. Meaning: for ANY (EVERY) wwfs P and Q, ((|- P) <=> (|- Q)) => |- (P <-> Q). Now you CAN'T prove a GENERAL statement with a special case (specific example). BUT you can disprove it with a _single_ COUNTEREXAMPLE. That's what David C. Ullrich did. >> >> I didn't say that it is never the case that P==Q is >> provable. I said that this does not follow [in general --F.] >> from the assumption that neither P nor Q is provable. >> > I was just illustrating my thinking with a specific example. > See comments above. > > A better counterexample for you (than meaningless propositional variables) > would <bla and bla> > Look man, this counterexample is perfectly "valid". W e do n o t need more (or others) than that. F. P.S. But |- P & |- Q => |- (P <-> Q) would hold. -- "I do tend to feel Hughes & Cresswell is a more authoritative source than you." (David C. Ullrich)
From: Charlie-Boo on 20 Dec 2005 11:08 H. J. Sander Bruggink wrote: > Charlie-Boo wrote: > > H. J. Sander Bruggink wrote: > > >>But you did talk about "truth tables" and "resolution", both > >>of which are proof methods for classical propositional logic, > >>and not intuitionistic propositional logic. > > > > Please give a proof of a propositional calculus proposition that cannot > > be done using case analysis (examination of the truth tables.) > > Man, just admit that you don't know what intuitionistic logic is. You can use any logic you want, actually. Please answer the question, ok? > You don't mention the word "intuitionistic" in your question > anymore, Never did. > making it ambiguous in the present context. > > If I interpret your question as: > > "Please give an intuitionistic proof of an > (intuitionistically valid) propositional calculus > proposition that cannot be done using case > analysis." > > then the answer would be: *any proof*, because "case analysis" > doesn't work for intuitionistic logic. One example will do. What is the propositional calculus wff that you prove? > groente > -- Sander
From: H. J. Sander Bruggink on 20 Dec 2005 11:27
Charlie-Boo wrote: > H. J. Sander Bruggink wrote: >>Man, just admit that you don't know what intuitionistic logic is. > > You can use any logic you want, actually. Please answer the question, > ok? You can't, and I did. >>You don't mention the word "intuitionistic" in your question >>anymore, > > Never did. Why are you posting in this subthread, which *is* about intuitionistic logic? (You DO know that sometimes subthreads digress from the original topic, don't you?) >>making it ambiguous in the present context. >> >>If I interpret your question as: >> >> "Please give an intuitionistic proof of an >> (intuitionistically valid) propositional calculus >> proposition that cannot be done using case >> analysis." >> >>then the answer would be: *any proof*, because "case analysis" >>doesn't work for intuitionistic logic. > > One example will do. What is the propositional calculus wff that you > prove? This seems futile, but ok, I'll bite. Here's an intuitionistic proof: 1. | P |---- 2. | P (rep) 3. P -> P (->I) Please show, by a "case analysis", that P->P is intuitionistically valid. groente -- Sander |