0.00/0.37 YES 0.00/0.37 0.00/0.37 DP problem for innermost termination. 0.00/0.37 P = 0.00/0.37 init#(x1) -> f1#(rnd1) 0.00/0.37 f2#(I0) -> f2#(I0 + 2) [I0 <= 19 /\ 10 <= I0 - 1] 0.00/0.37 f2#(I1) -> f2#(I1 + 2) [I1 <= 19 /\ I1 <= 10] 0.00/0.37 f1#(I2) -> f2#(0) 0.00/0.37 R = 0.00/0.37 init(x1) -> f1(rnd1) 0.00/0.37 f2(I0) -> f2(I0 + 2) [I0 <= 19 /\ 10 <= I0 - 1] 0.00/0.37 f2(I1) -> f2(I1 + 2) [I1 <= 19 /\ I1 <= 10] 0.00/0.37 f1(I2) -> f2(0) 0.00/0.37 0.00/0.37 The dependency graph for this problem is: 0.00/0.37 0 -> 3 0.00/0.37 1 -> 1 0.00/0.37 2 -> 1, 2 0.00/0.37 3 -> 2 0.00/0.37 Where: 0.00/0.37 0) init#(x1) -> f1#(rnd1) 0.00/0.37 1) f2#(I0) -> f2#(I0 + 2) [I0 <= 19 /\ 10 <= I0 - 1] 0.00/0.37 2) f2#(I1) -> f2#(I1 + 2) [I1 <= 19 /\ I1 <= 10] 0.00/0.37 3) f1#(I2) -> f2#(0) 0.00/0.37 0.00/0.37 We have the following SCCs. 0.00/0.37 { 2 } 0.00/0.37 { 1 } 0.00/0.37 0.00/0.37 DP problem for innermost termination. 0.00/0.37 P = 0.00/0.37 f2#(I0) -> f2#(I0 + 2) [I0 <= 19 /\ 10 <= I0 - 1] 0.00/0.37 R = 0.00/0.37 init(x1) -> f1(rnd1) 0.00/0.37 f2(I0) -> f2(I0 + 2) [I0 <= 19 /\ 10 <= I0 - 1] 0.00/0.37 f2(I1) -> f2(I1 + 2) [I1 <= 19 /\ I1 <= 10] 0.00/0.37 f1(I2) -> f2(0) 0.00/0.37 0.00/0.37 We use the reverse value criterion with the projection function NU: 0.00/0.37 NU[f2#(z1)] = 19 + -1 * z1 0.00/0.37 0.00/0.37 This gives the following inequalities: 0.00/0.37 I0 <= 19 /\ 10 <= I0 - 1 ==> 19 + -1 * I0 > 19 + -1 * (I0 + 2) with 19 + -1 * I0 >= 0 0.00/0.37 0.00/0.37 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/0.37 0.00/0.37 DP problem for innermost termination. 0.00/0.37 P = 0.00/0.37 f2#(I1) -> f2#(I1 + 2) [I1 <= 19 /\ I1 <= 10] 0.00/0.37 R = 0.00/0.37 init(x1) -> f1(rnd1) 0.00/0.37 f2(I0) -> f2(I0 + 2) [I0 <= 19 /\ 10 <= I0 - 1] 0.00/0.37 f2(I1) -> f2(I1 + 2) [I1 <= 19 /\ I1 <= 10] 0.00/0.37 f1(I2) -> f2(0) 0.00/0.37 0.00/0.37 We use the reverse value criterion with the projection function NU: 0.00/0.37 NU[f2#(z1)] = 10 + -1 * z1 0.00/0.37 0.00/0.37 This gives the following inequalities: 0.00/0.37 I1 <= 19 /\ I1 <= 10 ==> 10 + -1 * I1 > 10 + -1 * (I1 + 2) with 10 + -1 * I1 >= 0 0.00/0.37 0.00/0.37 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.35 EOF