Clarification, so: 1. (a -> b -> c -> d) U (a b c => d) = (a b c => d)
2. (a -> b) U (a b c => d) = _|_ 3. (a -> b) U (a -> b -> c -> d) = (a =>b -> c -> d) Keean. On 31 March 2015 at 16:32, Keean Schupke <[email protected]> wrote: > > Why isn't (f : a b c => d) valid? It unifies with (a->b->c->d), (and >> therefore (a->b)) doesn't it? > > > No, it only unifies with (a->b->c->d) not (a->b). > > f: a b c => d > > requires a minimum of 3 arguments > > > Keean. > >
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