MAYBE

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

Proof:
 Open