MAYBE

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

Proof:
 Open