MAYBE

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

Proof:
 Open