MAYBE

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

Proof:
 Open