MAYBE
0.08/0.19	MAYBE
0.08/0.19	
0.08/0.19	Problem:
0.08/0.19	 dbl(0()) -> 0()
0.08/0.19	 dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
0.08/0.19	 dbls(nil()) -> nil()
0.08/0.19	 dbls(cons(X,Y)) -> cons(n__dbl(activate(X)),n__dbls(activate(Y)))
0.08/0.19	 sel(0(),cons(X,Y)) -> activate(X)
0.08/0.19	 sel(s(X),cons(Y,Z)) -> sel(activate(X),activate(Z))
0.08/0.19	 indx(nil(),X) -> nil()
0.08/0.19	 indx(cons(X,Y),Z) -> cons(n__sel(activate(X),activate(Z)),n__indx(activate(Y),activate(Z)))
0.08/0.19	 from(X) -> cons(activate(X),n__from(n__s(activate(X))))
0.08/0.19	 s(X) -> n__s(X)
0.08/0.19	 dbl(X) -> n__dbl(X)
0.08/0.19	 dbls(X) -> n__dbls(X)
0.08/0.19	 sel(X1,X2) -> n__sel(X1,X2)
0.08/0.19	 indx(X1,X2) -> n__indx(X1,X2)
0.08/0.19	 from(X) -> n__from(X)
0.08/0.19	 activate(n__s(X)) -> s(X)
0.08/0.19	 activate(n__dbl(X)) -> dbl(X)
0.08/0.19	 activate(n__dbls(X)) -> dbls(X)
0.08/0.19	 activate(n__sel(X1,X2)) -> sel(X1,X2)
0.08/0.19	 activate(n__indx(X1,X2)) -> indx(X1,X2)
0.08/0.19	 activate(n__from(X)) -> from(X)
0.08/0.19	 activate(X) -> X
0.08/0.19	
0.08/0.19	Proof:
0.08/0.19	 Open
0.08/0.19	EOF