R J <rj248...@hotmail.com> writes: > I'm trying to prove that (==) is reflexive, symmetric, and > transitive over the Bools, given this definition:
> (==):: Bool -> Bool -> Bool > x == y = (x && y) || (not x && not y) Since Bool is a type, and all Haskell types include ⊥, you need to add conditions in your proofs to exclude it. -- Jón Fairbairn jon.fairba...@cl.cam.ac.uk _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
