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 f3#(I0) -> f3#(I0 + 3) [I0 <= 20] 0.00/0.37 f2#(I1) -> f3#(5) [99 <= I1 - 1] 0.00/0.37 f2#(I2) -> f2#(I2 + 1) [I2 <= 99] 0.00/0.37 f1#(I3) -> f2#(0) 0.00/0.37 R = 0.00/0.37 init(x1) -> f1(rnd1) 0.00/0.37 f3(I0) -> f3(I0 + 3) [I0 <= 20] 0.00/0.37 f2(I1) -> f3(5) [99 <= I1 - 1] 0.00/0.37 f2(I2) -> f2(I2 + 1) [I2 <= 99] 0.00/0.37 f1(I3) -> f2(0) 0.00/0.37 0.00/0.37 The dependency graph for this problem is: 0.00/0.37 0 -> 4 0.00/0.37 1 -> 1 0.00/0.37 2 -> 1 0.00/0.37 3 -> 2, 3 0.00/0.37 4 -> 3 0.00/0.37 Where: 0.00/0.37 0) init#(x1) -> f1#(rnd1) 0.00/0.37 1) f3#(I0) -> f3#(I0 + 3) [I0 <= 20] 0.00/0.37 2) f2#(I1) -> f3#(5) [99 <= I1 - 1] 0.00/0.37 3) f2#(I2) -> f2#(I2 + 1) [I2 <= 99] 0.00/0.37 4) f1#(I3) -> f2#(0) 0.00/0.37 0.00/0.37 We have the following SCCs. 0.00/0.37 { 3 } 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 f3#(I0) -> f3#(I0 + 3) [I0 <= 20] 0.00/0.37 R = 0.00/0.37 init(x1) -> f1(rnd1) 0.00/0.37 f3(I0) -> f3(I0 + 3) [I0 <= 20] 0.00/0.37 f2(I1) -> f3(5) [99 <= I1 - 1] 0.00/0.37 f2(I2) -> f2(I2 + 1) [I2 <= 99] 0.00/0.37 f1(I3) -> 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[f3#(z1)] = 20 + -1 * z1 0.00/0.37 0.00/0.37 This gives the following inequalities: 0.00/0.37 I0 <= 20 ==> 20 + -1 * I0 > 20 + -1 * (I0 + 3) with 20 + -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#(I2) -> f2#(I2 + 1) [I2 <= 99] 0.00/0.37 R = 0.00/0.37 init(x1) -> f1(rnd1) 0.00/0.37 f3(I0) -> f3(I0 + 3) [I0 <= 20] 0.00/0.37 f2(I1) -> f3(5) [99 <= I1 - 1] 0.00/0.37 f2(I2) -> f2(I2 + 1) [I2 <= 99] 0.00/0.37 f1(I3) -> 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)] = 99 + -1 * z1 0.00/0.37 0.00/0.37 This gives the following inequalities: 0.00/0.37 I2 <= 99 ==> 99 + -1 * I2 > 99 + -1 * (I2 + 1) with 99 + -1 * I2 >= 0 0.00/0.37 0.00/0.37 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.36 EOF