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stiff

Reductio ad absurdum is:

(x->(~x))->(~x)

Law of excluded middle is:

x | ~x

Proof follows:

(x->(~x))->(~x)

From definition of implication is equivalent to:

(~x) | ~(x->(~x))

From definition of implication is equivalent to:

(~x) | ~((~x) | (~x))

From de Morgans law is equivalent to:

(~x) | (x & x)

From identity law is equivalent to:

~x | x

I am not a logician and I might not have gotten all the details right, but I certainly think this is possible. I also have quite a good proof by authority ;) in form of an article by Alonzo Church:

http://www.ams.org/journals/bull/1928-34-01/S0002-9904-1928-...

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That looks correct to me.

Actually, at first I thought it was incorrect because you used De Morgan's law, which I mistakenly thought was invalid in intuitionistic logic. However, I looked it up and apparently only ~(p & q) |- ~p v ~q is invalid in IL.

Thanks, that was the part I felt uncertain about, funny how this seems completely elementary on one hand and on the other it is so easy to make a mistake.

Can you justify using material implication?

It's possible to create reductio ad absurdum in Coq:

    Definition reductioAdAbsurdum (X:Prop) (f : (X -> (X -> False))) (x:X) : False := f x x.

But not the law of the excluded middle.

Edit: It's much clearer in Idris:

    reductioAdAbsurdum : (x -> (x -> _|_)) -> (x -> _|_)
    reductioAdAbsurdum f x = f x x

I am assuming that we are talking about taking the law of excluded middle out from classical logic, where all the other things independent from that law still hold, and I think implication being material is independent.

I mean material implication as in:

In propositional logic, material implication is a valid rule of replacement which is an instance of the connective of the same name. It is the rule that states that "P implies Q" is logically equivalent to "not-P or Q".

http://en.wikipedia.org/wiki/Implication

Yes I understand, but why do you think it needs justification? As far as I understand this is simply the definition of implication in classical logic, for one it follows from the truth tables of both expressions.

OK, but we're not talking about classical logic, otherwise the question about whether the law of the excluded middle is derivable is uninteresting.

Well, that's exactly what people were considering and what the article is referring too, a variant of classical logic where everything except the law of excluded middle holds true. I don't think this is uninteresting, which postulates are independent from one another is one of the core problems in logic.

Nice flame, eot.

I didn't mean to flame, I thought we were having an interesting debate. I was under the impression we we're talking about logic systems in general where reductio ad absurdum exists, but the law of the excluded middle doesn't. Anyway, thanks, I learnt much.

This would be a very peculiar logic I think, maybe you just come from a very different background and hence all the hair-splitting was necessary, we just wandered so far away from the original point I lost patience, maybe unnecessarily, in the end it really was an interesting debate for me too.

Actually, I guess he's right.

Arguably, the common sense definition of 'A -> B' is simply modus ponens. That is 'Given A and A->B, it follows that B'. It's then no longer obvious that A -> B is equivalent to ~A v B.

Indeed, implication in IL is different from material implication even though IL has modus ponens.

Actually, this is wrong. Reductio ad absurdum should be:

    ((~x) -> x) -> x

or:

    ((x -> _|_) -> x) -> x

Which is not constructible in Coq/Idris.

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