MAYBE

Problem:
 f(0(x1)) -> s(0(x1))
 d(0(x1)) -> 0(x1)
 d(s(x1)) -> s(s(d(p(s(x1)))))
 f(s(x1)) -> d(f(p(s(x1))))
 p(s(x1)) -> x1

Proof:
 Open