MAYBE

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

Proof:
 Open