MAYBE

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

Proof:
 Open