0.74/0.97 MAYBE 0.74/0.97 0.74/0.97 DP problem for innermost termination. 0.74/0.97 P = 0.74/0.97 eval#(x, y, z) -> eval#(x, y - 1, z + y) [x >= 0 && z * z * z >= y] 0.74/0.97 eval#(I0, I1, I2) -> eval#(I0 - 1, I1 - 1, I2) [I0 >= 0 && I2 * I2 * I2 >= I1] 0.74/0.97 R = 0.74/0.97 eval(x, y, z) -> eval(x, y - 1, z + y) [x >= 0 && z * z * z >= y] 0.74/0.97 eval(I0, I1, I2) -> eval(I0 - 1, I1 - 1, I2) [I0 >= 0 && I2 * I2 * I2 >= I1] 0.74/0.97 0.74/0.97 We use the reverse value criterion with the projection function NU: 0.74/0.97 NU[eval#(z1,z2,z3)] = z1 0.74/0.97 0.74/0.97 This gives the following inequalities: 0.74/0.97 x >= 0 && z * z * z >= y ==> x >= x 0.74/0.97 I0 >= 0 && I2 * I2 * I2 >= I1 ==> I0 > I0 - 1 with I0 >= 0 0.74/0.97 0.74/0.97 We remove all the strictly oriented dependency pairs. 0.74/0.97 0.74/0.97 DP problem for innermost termination. 0.74/0.97 P = 0.74/0.97 eval#(x, y, z) -> eval#(x, y - 1, z + y) [x >= 0 && z * z * z >= y] 0.74/0.97 R = 0.74/0.97 eval(x, y, z) -> eval(x, y - 1, z + y) [x >= 0 && z * z * z >= y] 0.74/0.97 eval(I0, I1, I2) -> eval(I0 - 1, I1 - 1, I2) [I0 >= 0 && I2 * I2 * I2 >= I1] 0.74/0.97 0.74/3.95 EOF