R J <[email protected]> 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 [email protected] _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
