0.00/0.34 YES 0.00/0.34 0.00/0.34 DP problem for innermost termination. 0.00/0.34 P = 0.00/0.34 eval_1#(x, y, z) -> eval_1#(x, y, x - y) [y > x && z > y && y > 0] 0.00/0.34 eval_1#(I0, I1, I2) -> eval_1#(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0] 0.00/0.34 eval_0#(I3, I4, I5) -> eval_1#(I3, I4, I5) [I4 > 0] 0.00/0.34 R = 0.00/0.34 eval_1(x, y, z) -> eval_1(x, y, x - y) [y > x && z > y && y > 0] 0.00/0.34 eval_1(I0, I1, I2) -> eval_1(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0] 0.00/0.34 eval_0(I3, I4, I5) -> eval_1(I3, I4, I5) [I4 > 0] 0.00/0.34 0.00/0.34 The dependency graph for this problem is: 0.00/0.34 0 -> 0.00/0.34 1 -> 0, 1 0.00/0.34 2 -> 0, 1 0.00/0.34 Where: 0.00/0.34 0) eval_1#(x, y, z) -> eval_1#(x, y, x - y) [y > x && z > y && y > 0] 0.00/0.34 1) eval_1#(I0, I1, I2) -> eval_1#(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0] 0.00/0.34 2) eval_0#(I3, I4, I5) -> eval_1#(I3, I4, I5) [I4 > 0] 0.00/0.34 0.00/0.34 We have the following SCCs. 0.00/0.34 { 1 } 0.00/0.34 0.00/0.34 DP problem for innermost termination. 0.00/0.34 P = 0.00/0.34 eval_1#(I0, I1, I2) -> eval_1#(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0] 0.00/0.34 R = 0.00/0.34 eval_1(x, y, z) -> eval_1(x, y, x - y) [y > x && z > y && y > 0] 0.00/0.34 eval_1(I0, I1, I2) -> eval_1(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0] 0.00/0.34 eval_0(I3, I4, I5) -> eval_1(I3, I4, I5) [I4 > 0] 0.00/0.34 0.00/0.34 We use the reverse value criterion with the projection function NU: 0.00/0.34 NU[eval_1#(z1,z2,z3)] = z2 + -1 * z1 0.00/0.34 0.00/0.34 This gives the following inequalities: 0.00/0.34 I1 > I0 && I2 > I1 && I1 > 0 ==> I1 + -1 * I0 > I1 + -1 * (I0 + I1) with I1 + -1 * I0 >= 0 0.00/0.34 0.00/0.34 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.32 EOF