0.00/0.03 YES 0.00/0.03 0.00/0.03 DP problem for innermost termination. 0.00/0.03 P = 0.00/0.03 g#(x, cons(y, ys)) -> g#(x, ys) 0.00/0.03 R = 0.00/0.03 g(x, cons(y, ys)) -> cons(x + y, g(x, ys)) 0.00/0.03 0.00/0.03 We use the subterm criterion with the projection function NU: 0.00/0.03 NU[g#(x0,x1)] = x1 0.00/0.03 0.00/0.03 This gives the following inequalities: 0.00/0.03 cons(y, ys) |> ys 0.00/0.03 0.00/0.03 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.01 EOF