[Haskell-cafe] lazy evaluation

Can anybody give me a simple explanation why the second definition of a palindrome checker does not terminate, although the first one does? pal :: Eq a => [a] -> Bool pal s = b where (b,r) = eqrev s r [] eqrev :: Eq a => [a] -> [a] -> [a] -> (Bool,[a]) eqrev (x:s1) ~(y:s2) acc = (x==y&&b,r) where (b,r) = eqrev s1 s2 (x:acc) eqrev _ _ acc = (True,acc) pal :: Eq a => [a] -> Bool pal s = b where (b,r) = eqrev' s r eqrev' :: Eq a => [a] -> [a] -> (Bool,[a]) eqrev' (x:s1) ~(y:s2) = (x==y&&b,r++[y]) where (b,r) = eqrev' s1 s2 eqrev' _ _ = (True,[]) Peter