MAYBE

Problem:
 b(b(x1)) -> b(a(b(a(a(b(x1))))))
 b(b(a(b(x1)))) -> b(a(b(b(x1))))
 b(a(a(a(b(a(a(b(x1)))))))) -> b(a(a(a(b(b(x1))))))
 b(b(a(a(b(x1))))) -> b(a(a(b(b(x1)))))

Proof:
 Open