Peter Padawitz writes: > Is f(~p(x))=e(x) semantically equivalent to: f(z)=e(x) where p(x)=z?
Yep. See also http://en.wikibooks.org/wiki/Haskell/Laziness#Lazy_pattern_matching regarding lazy patterns. -- -David House, [EMAIL PROTECTED] _______________________________________________ Haskell mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell
