In classical logic A -> B is the equivalent to ~A v B (with ~ = not and v = or)
So (forall a. P(a)) -> Q {implication = not-or} ~(forall a. P(a)) v Q {forall a. X is equivalent to there does not exist a such that X doesn't hold} ~(~exists a. ~P(a)) v Q {double negation elimination} (exists a. ~P(a)) v Q {a is not free in Q} exists a. (~P(a) v Q) {implication = not-or} exists a. (P(a) -> Q) These steps are all equivalencies, valid in both directions. On Wed, Aug 15, 2012 at 9:32 AM, David Feuer <david.fe...@gmail.com> wrote: > On Aug 15, 2012 3:21 AM, "wren ng thornton" <w...@freegeek.org> wrote: > > It's even easier than that. > > > > (forall a. P(a)) -> Q <=> exists a. (P(a) -> Q) > > > > Where P and Q are metatheoretic/schematic variables. This is just the > usual thing about antecedents being in a "negative" position, and thus > flipping as you move into/out of that position. > > Most of this conversation is going over my head. I can certainly see how > exists a. (P(a)->Q) implies that (forall a. P(a))->Q. The opposite > certainly doesn't hold in classical logic. What sort of logic are you folks > working in? > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > >
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