MAYBE

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

Proof:
 Open