MAYBE

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

Proof:
 Open