MAYBE
0.09/0.17	MAYBE
0.09/0.20	
0.09/0.20	Problem:
0.09/0.20	 g(X) -> u(h(X),h(X),X)
0.09/0.20	 u(d(),c(Y),X) -> k(Y)
0.09/0.20	 h(d()) -> c(a())
0.09/0.20	 h(d()) -> c(b())
0.09/0.20	 f(k(a()),k(b()),X) -> f(X,X,X)
0.09/0.20	
0.09/0.20	Proof:
0.09/0.20	 Open
0.09/0.20	EOF