0.00/0.26 YES 0.00/0.26 0.00/0.26 DP problem for innermost termination. 0.00/0.26 P = 0.00/0.26 init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.26 f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] 0.00/0.26 f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1] 0.00/0.26 R = 0.00/0.26 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.26 f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] 0.00/0.26 f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1] 0.00/0.26 0.00/0.26 The dependency graph for this problem is: 0.00/0.26 0 -> 2 0.00/0.26 1 -> 1 0.00/0.26 2 -> 1 0.00/0.26 Where: 0.00/0.26 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.26 1) f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] 0.00/0.26 2) f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1] 0.00/0.26 0.00/0.26 We have the following SCCs. 0.00/0.26 { 1 } 0.00/0.26 0.00/0.26 DP problem for innermost termination. 0.00/0.26 P = 0.00/0.26 f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] 0.00/0.26 R = 0.00/0.26 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.26 f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] 0.00/0.26 f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1] 0.00/0.26 0.00/0.26 We use the reverse value criterion with the projection function NU: 0.00/0.26 NU[f2#(z1,z2)] = z1 - 1 + -1 * z2 0.00/0.26 0.00/0.26 This gives the following inequalities: 0.00/0.26 I1 <= I0 - 1 ==> I0 - 1 + -1 * I1 > I0 - 1 + -1 * (I1 + 1) with I0 - 1 + -1 * I1 >= 0 0.00/0.26 0.00/0.26 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.24 EOF