MAYBE

Problem:
 f(x,y) -> cond(and(isNat(x),isNat(y)),x,y)
 cond(tt(),x,y) -> f(s(x),s(y))
 isNat(s(x)) -> isNat(x)
 isNat(0()) -> tt()
 and(tt(),tt()) -> tt()
 and(ff(),x) -> ff()
 and(x,ff()) -> ff()

Proof:
 Open