MAYBE
0.08/0.18	MAYBE
0.08/0.18	
0.08/0.18	Problem:
0.08/0.18	 f(X,n__g(X),Y) -> f(activate(Y),activate(Y),activate(Y))
0.08/0.18	 g(b()) -> c()
0.08/0.18	 b() -> c()
0.08/0.18	 g(X) -> n__g(X)
0.08/0.18	 activate(n__g(X)) -> g(activate(X))
0.08/0.18	 activate(X) -> X
0.08/0.18	
0.08/0.18	Proof:
0.08/0.18	 Open
0.08/0.18	EOF