MAYBE

Problem:
 a(b(b(x1))) -> P(a(b(x1)))
 a(P(x1)) -> P(a(x(x1)))
 a(x(x1)) -> x(a(x1))
 b(P(x1)) -> b(Q(x1))
 Q(x(x1)) -> a(Q(x1))
 Q(a(x1)) -> b(b(a(x1)))

Proof:
 Open