MAYBE

Problem:
 g(h(x1)) -> g(f(s(x1)))
 f(s(s(s(x1)))) -> h(f(s(h(x1))))
 f(h(x1)) -> h(f(s(h(x1))))
 h(x1) -> x1
 f(f(s(s(x1)))) -> s(s(s(f(f(x1)))))
 b(a(x1)) -> a(b(x1))
 a(a(a(x1))) -> b(a(a(b(x1))))
 b(b(b(b(x1)))) -> a(x1)

Proof:
 Open