0.00/0.12 YES 0.00/0.12 0.00/0.12 DP problem for innermost termination. 0.00/0.12 P = 0.00/0.12 f4#(x1) -> f1#(x1) 0.00/0.12 f3#(I0) -> f2#(I0) 0.00/0.12 f2#(I1) -> f3#(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] 0.00/0.12 f1#(I2) -> f2#(I2) 0.00/0.12 R = 0.00/0.12 f4(x1) -> f1(x1) 0.00/0.12 f3(I0) -> f2(I0) 0.00/0.12 f2(I1) -> f3(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] 0.00/0.12 f1(I2) -> f2(I2) 0.00/0.12 0.00/0.12 The dependency graph for this problem is: 0.00/0.12 0 -> 3 0.00/0.12 1 -> 2 0.00/0.12 2 -> 1 0.00/0.12 3 -> 2 0.00/0.12 Where: 0.00/0.12 0) f4#(x1) -> f1#(x1) 0.00/0.12 1) f3#(I0) -> f2#(I0) 0.00/0.12 2) f2#(I1) -> f3#(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] 0.00/0.12 3) f1#(I2) -> f2#(I2) 0.00/0.12 0.00/0.12 We have the following SCCs. 0.00/0.12 { 1, 2 } 0.00/0.12 0.00/0.12 DP problem for innermost termination. 0.00/0.12 P = 0.00/0.12 f3#(I0) -> f2#(I0) 0.00/0.12 f2#(I1) -> f3#(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] 0.00/0.12 R = 0.00/0.12 f4(x1) -> f1(x1) 0.00/0.12 f3(I0) -> f2(I0) 0.00/0.12 f2(I1) -> f3(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] 0.00/0.12 f1(I2) -> f2(I2) 0.00/0.12 0.00/0.12 We use the basic value criterion with the projection function NU: 0.00/0.12 NU[f2#(z1)] = z1 0.00/0.12 NU[f3#(z1)] = z1 0.00/0.12 0.00/0.12 This gives the following inequalities: 0.00/0.12 ==> I0 (>! \union =) I0 0.00/0.12 0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1 ==> I1 >! rnd1 0.00/0.12 0.00/0.12 We remove all the strictly oriented dependency pairs. 0.00/0.12 0.00/0.12 DP problem for innermost termination. 0.00/0.12 P = 0.00/0.12 f3#(I0) -> f2#(I0) 0.00/0.12 R = 0.00/0.12 f4(x1) -> f1(x1) 0.00/0.12 f3(I0) -> f2(I0) 0.00/0.12 f2(I1) -> f3(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] 0.00/0.12 f1(I2) -> f2(I2) 0.00/0.12 0.00/0.12 The dependency graph for this problem is: 0.00/0.12 1 -> 0.00/0.12 Where: 0.00/0.12 1) f3#(I0) -> f2#(I0) 0.00/0.12 0.00/0.12 We have the following SCCs. 0.00/0.12 0.00/3.10 EOF