g1 0 []     = 0
g1 _ (x:xs) = x

g2 _ (x:xs) = x
g2 0 []     = 0

h x = h x

-- The evaluation of `m1` does not terminate in Haskell:
m1 = g1 (h 0) [1]

-- The evaluation of `m2` terminates in Haskell:
m2 = g2 (h 0) [1]

-- > Consequence:
-- > All rules are equal but some rules are more equal
-- > (George Orwell: Animal Farm, adapted to Haskell)