0.00/0.14 YES 0.00/0.14 0.00/0.14 DP problem for innermost termination. 0.00/0.14 P = 0.00/0.14 eval#(x, y, z) -> eval#(x, y + 1, z + 1) [x > y + z] 0.00/0.14 R = 0.00/0.14 eval(x, y, z) -> eval(x, y + 1, z + 1) [x > y + z] 0.00/0.14 0.00/0.14 We use the reverse value criterion with the projection function NU: 0.00/0.14 NU[eval#(z1,z2,z3)] = z1 + -1 * (z2 + z3) 0.00/0.14 0.00/0.14 This gives the following inequalities: 0.00/0.14 x > y + z ==> x + -1 * (y + z) > x + -1 * (y + 1 + (z + 1)) with x + -1 * (y + z) >= 0 0.00/0.14 0.00/0.14 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.12 EOF