0.00/0.21 YES 0.00/0.21 0.00/0.21 DP problem for innermost termination. 0.00/0.21 P = 0.00/0.21 init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.21 f2#(I0, I1) -> f2#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] 0.00/0.21 f1#(I4, I5) -> f2#(I6, I7) [3 <= I6 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I6 - 3 <= I4] 0.00/0.21 R = 0.00/0.21 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.21 f2(I0, I1) -> f2(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] 0.00/0.21 f1(I4, I5) -> f2(I6, I7) [3 <= I6 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I6 - 3 <= I4] 0.00/0.21 0.00/0.21 The dependency graph for this problem is: 0.00/0.21 0 -> 2 0.00/0.21 1 -> 1 0.00/0.21 2 -> 1 0.00/0.21 Where: 0.00/0.21 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.21 1) f2#(I0, I1) -> f2#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] 0.00/0.21 2) f1#(I4, I5) -> f2#(I6, I7) [3 <= I6 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I6 - 3 <= I4] 0.00/0.21 0.00/0.21 We have the following SCCs. 0.00/0.21 { 1 } 0.00/0.21 0.00/0.21 DP problem for innermost termination. 0.00/0.21 P = 0.00/0.21 f2#(I0, I1) -> f2#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] 0.00/0.21 R = 0.00/0.21 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.21 f2(I0, I1) -> f2(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] 0.00/0.21 f1(I4, I5) -> f2(I6, I7) [3 <= I6 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I6 - 3 <= I4] 0.00/0.21 0.00/0.21 We use the basic value criterion with the projection function NU: 0.00/0.21 NU[f2#(z1,z2)] = z1 0.00/0.21 0.00/0.21 This gives the following inequalities: 0.00/0.21 -1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0 ==> I0 >! I2 0.00/0.21 0.00/0.21 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.19 EOF