MAYBE

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

Proof:
 Open