MAYBE

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

Proof:
 Open