MAYBE Problem: b(b(x)) -> a(c(x)) c(a(x)) -> a(a(x)) b(a(x)) -> a(c(x)) b(c(x)) -> b(b(x)) a(a(x)) -> b(b(x)) c(c(x)) -> b(c(x)) a(b(x)) -> a(c(x)) Proof: Open