MAYBE

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

Proof:
 Open