MAYBE
0.08/0.18	MAYBE
0.08/0.18	
0.08/0.18	Problem:
0.08/0.18	 p(0()) -> 0()
0.08/0.18	 p(s(X)) -> X
0.08/0.18	 leq(0(),Y) -> true()
0.08/0.18	 leq(s(X),0()) -> false()
0.08/0.18	 leq(s(X),s(Y)) -> leq(X,Y)
0.08/0.18	 if(true(),X,Y) -> activate(X)
0.08/0.18	 if(false(),X,Y) -> activate(Y)
0.08/0.18	 diff(X,Y) -> if(leq(X,Y),n__0(),n__s(n__diff(n__p(X),Y)))
0.08/0.18	 0() -> n__0()
0.08/0.18	 s(X) -> n__s(X)
0.08/0.18	 diff(X1,X2) -> n__diff(X1,X2)
0.08/0.18	 p(X) -> n__p(X)
0.08/0.18	 activate(n__0()) -> 0()
0.08/0.18	 activate(n__s(X)) -> s(activate(X))
0.08/0.18	 activate(n__diff(X1,X2)) -> diff(activate(X1),activate(X2))
0.08/0.18	 activate(n__p(X)) -> p(activate(X))
0.08/0.18	 activate(X) -> X
0.08/0.18	
0.08/0.18	Proof:
0.08/0.18	 Open
0.08/0.19	EOF