0.00/0.11 YES 0.00/0.11 0.00/0.11 DP problem for innermost termination. 0.00/0.11 P = 0.00/0.11 f4#(x1) -> f3#(x1) 0.00/0.11 f3#(I0) -> f1#(I0) 0.00/0.11 f2#(I1) -> f1#(I1) 0.00/0.11 f1#(I2) -> f2#(-1000 + I2) [1 <= -1000 + I2] 0.00/0.11 R = 0.00/0.11 f4(x1) -> f3(x1) 0.00/0.11 f3(I0) -> f1(I0) 0.00/0.11 f2(I1) -> f1(I1) 0.00/0.11 f1(I2) -> f2(-1000 + I2) [1 <= -1000 + I2] 0.00/0.11 0.00/0.11 The dependency graph for this problem is: 0.00/0.11 0 -> 1 0.00/0.11 1 -> 3 0.00/0.11 2 -> 3 0.00/0.11 3 -> 2 0.00/0.11 Where: 0.00/0.11 0) f4#(x1) -> f3#(x1) 0.00/0.11 1) f3#(I0) -> f1#(I0) 0.00/0.11 2) f2#(I1) -> f1#(I1) 0.00/0.11 3) f1#(I2) -> f2#(-1000 + I2) [1 <= -1000 + I2] 0.00/0.11 0.00/0.11 We have the following SCCs. 0.00/0.11 { 2, 3 } 0.00/0.11 0.00/0.11 DP problem for innermost termination. 0.00/0.11 P = 0.00/0.11 f2#(I1) -> f1#(I1) 0.00/0.11 f1#(I2) -> f2#(-1000 + I2) [1 <= -1000 + I2] 0.00/0.11 R = 0.00/0.11 f4(x1) -> f3(x1) 0.00/0.11 f3(I0) -> f1(I0) 0.00/0.11 f2(I1) -> f1(I1) 0.00/0.11 f1(I2) -> f2(-1000 + I2) [1 <= -1000 + I2] 0.00/0.11 0.00/0.11 We use the basic value criterion with the projection function NU: 0.00/0.11 NU[f1#(z1)] = z1 0.00/0.11 NU[f2#(z1)] = z1 0.00/0.11 0.00/0.11 This gives the following inequalities: 0.00/0.11 ==> I1 (>! \union =) I1 0.00/0.11 1 <= -1000 + I2 ==> I2 >! -1000 + I2 0.00/0.11 0.00/0.11 We remove all the strictly oriented dependency pairs. 0.00/0.11 0.00/0.11 DP problem for innermost termination. 0.00/0.11 P = 0.00/0.11 f2#(I1) -> f1#(I1) 0.00/0.11 R = 0.00/0.11 f4(x1) -> f3(x1) 0.00/0.11 f3(I0) -> f1(I0) 0.00/0.11 f2(I1) -> f1(I1) 0.00/0.11 f1(I2) -> f2(-1000 + I2) [1 <= -1000 + I2] 0.00/0.11 0.00/0.11 The dependency graph for this problem is: 0.00/0.11 2 -> 0.00/0.11 Where: 0.00/0.11 2) f2#(I1) -> f1#(I1) 0.00/0.11 0.00/0.11 We have the following SCCs. 0.00/0.11 0.00/3.09 EOF