[FOM] question about Mendelson's Intro to Math Logic
Paulo Oliva
p.oliva at qmul.ac.uk
Wed Feb 12 06:26:46 EST 2014
Hi Bob
I think you mean replacing it with (not B -> not C) -> (C -> B), the one you wrote (not B -> not C) -> (B -> C) is not valid.
Anyway, here is a sketch of how it should go. Assume
(*) (not B -> not C) -> (C -> B)
and let’s derive A3. If you take C be true (e.g. axiom A1) in (*) one can conclude not not B -> B, i.e. double negation elimination DNE.
Then, in order to derive (not C -> not B) -> (not C -> B) -> C one first assumes not C and given the two premises concludes B and not B,
and hence not not C. By DNE we get C.
We’ve used contraction (used not C twice) but that follows from A2.
Formalising all this just using cut will probably be painful, but it’s clear it can be done since there is
a simple translation from natural deduction to Hilbert style proofs.
On 10 Feb 2014, at 00:58, Robert Lubarsky <lubarsky.robert at comcast.net> wrote:
> There’s an exercise from Mendelson’s Intro to Mathematical Logic that I cannot do, and would like an answer to. In the 5th edition, it’s 1.59 on p. 39. Namely, he gives three axioms for a Hilbert-style propositional proof system. The first two are essentially the combinators s and k. The third is
> (not C -> not B) -> [ (not C -> B) -> C ]. The exercise in question is to show that, if you replace this last schema with (not B -> not C) -> (B -> C) then the new system proves the same as the old. I don’t see how to do this.
>
> My attempt is as follows. We can use the deduction theorem. So assume (not C -> not B) and (not C -> B), with the goal of proving C. From the new schema, we get B -> C. We can compose that with (not C -> B) to get (not C -> C). I don’t see how to go from that to C within this system.
>
> Bob Lubarsky
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--
Paulo Oliva
Royal Society University Research Fellow
Queen Mary University of London
London E1 4FZ
+44 (0) 207 882 5255
http://www.dcs.qmul.ac.uk/~pbo
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