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 f3#(x1) -> f1#(x1) 0.00/0.11 f2#(I0) -> f1#(I0) 0.00/0.11 f1#(I1) -> f2#(-1 + I1) [1 <= I1] 0.00/0.11 R = 0.00/0.11 f3(x1) -> f1(x1) 0.00/0.11 f2(I0) -> f1(I0) 0.00/0.11 f1(I1) -> f2(-1 + I1) [1 <= I1] 0.00/0.11 0.00/0.11 The dependency graph for this problem is: 0.00/0.11 0 -> 2 0.00/0.11 1 -> 2 0.00/0.11 2 -> 1 0.00/0.11 Where: 0.00/0.11 0) f3#(x1) -> f1#(x1) 0.00/0.11 1) f2#(I0) -> f1#(I0) 0.00/0.11 2) f1#(I1) -> f2#(-1 + I1) [1 <= I1] 0.00/0.11 0.00/0.11 We have the following SCCs. 0.00/0.11 { 1, 2 } 0.00/0.11 0.00/0.11 DP problem for innermost termination. 0.00/0.11 P = 0.00/0.11 f2#(I0) -> f1#(I0) 0.00/0.11 f1#(I1) -> f2#(-1 + I1) [1 <= I1] 0.00/0.11 R = 0.00/0.11 f3(x1) -> f1(x1) 0.00/0.11 f2(I0) -> f1(I0) 0.00/0.11 f1(I1) -> f2(-1 + I1) [1 <= I1] 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 ==> I0 (>! \union =) I0 0.00/0.11 1 <= I1 ==> I1 >! -1 + I1 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#(I0) -> f1#(I0) 0.00/0.11 R = 0.00/0.11 f3(x1) -> f1(x1) 0.00/0.11 f2(I0) -> f1(I0) 0.00/0.11 f1(I1) -> f2(-1 + I1) [1 <= I1] 0.00/0.11 0.00/0.11 The dependency graph for this problem is: 0.00/0.11 1 -> 0.00/0.11 Where: 0.00/0.11 1) f2#(I0) -> f1#(I0) 0.00/0.11 0.00/0.11 We have the following SCCs. 0.00/0.11 0.00/3.09 EOF