MAYBE

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

Proof:
 Open