MAYBE

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

Proof:
 Open