4.44/4.45 YES 4.44/4.45 4.44/4.45 DP problem for innermost termination. 4.44/4.45 P = 4.44/4.45 init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 4.44/4.45 f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3#(I3, I4, I5) -> f2#(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2#(I9, I10, I11) -> f3#(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2#(I12, I13, I14) -> f3#(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1#(I15, I16, I17) -> f2#(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 R = 4.44/4.45 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 4.44/4.45 f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 The dependency graph for this problem is: 4.44/4.45 0 -> 6 4.44/4.45 1 -> 4 4.44/4.45 2 -> 4.44/4.45 3 -> 4, 5 4.44/4.45 4 -> 1 4.44/4.45 5 -> 3 4.44/4.45 6 -> 4, 5 4.44/4.45 Where: 4.44/4.45 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 4.44/4.45 1) f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 2) f3#(I3, I4, I5) -> f2#(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 3) f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 4) f2#(I9, I10, I11) -> f3#(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 5) f2#(I12, I13, I14) -> f3#(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 6) f1#(I15, I16, I17) -> f2#(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 We have the following SCCs. 4.44/4.45 { 3, 5 } 4.44/4.45 { 1, 4 } 4.44/4.45 4.44/4.45 DP problem for innermost termination. 4.44/4.45 P = 4.44/4.45 f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f2#(I9, I10, I11) -> f3#(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 R = 4.44/4.45 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 4.44/4.45 f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 We use the reverse value criterion with the projection function NU: 4.44/4.45 NU[f2#(z1,z2,z3)] = z1 - 1 + -1 * z3 4.44/4.45 NU[f3#(z1,z2,z3)] = z1 - 1 - 1 + -1 * z3 4.44/4.45 4.44/4.45 This gives the following inequalities: 4.44/4.45 I2 <= I0 - 1 /\ I1 <= I2 ==> I0 - 1 - 1 + -1 * I2 >= I0 - 1 - 1 + -1 * I2 4.44/4.45 I11 <= I9 - 1 /\ I10 <= I11 ==> I9 - 1 + -1 * I11 > I9 - 1 - 1 + -1 * I11 with I9 - 1 + -1 * I11 >= 0 4.44/4.45 4.44/4.45 We remove all the strictly oriented dependency pairs. 4.44/4.45 4.44/4.45 DP problem for innermost termination. 4.44/4.45 P = 4.44/4.45 f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 R = 4.44/4.45 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 4.44/4.45 f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 The dependency graph for this problem is: 4.44/4.45 1 -> 4.44/4.45 Where: 4.44/4.45 1) f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 4.44/4.45 We have the following SCCs. 4.44/4.45 4.44/4.45 4.44/4.45 DP problem for innermost termination. 4.44/4.45 P = 4.44/4.45 f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2#(I12, I13, I14) -> f3#(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 R = 4.44/4.45 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 4.44/4.45 f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 We use the reverse value criterion with the projection function NU: 4.44/4.45 NU[f2#(z1,z2,z3)] = z2 - 1 + -1 * z3 4.44/4.45 NU[f3#(z1,z2,z3)] = z2 - 1 - 1 + -1 * z3 4.44/4.45 4.44/4.45 This gives the following inequalities: 4.44/4.45 I8 <= I7 - 1 ==> I7 - 1 - 1 + -1 * I8 >= I7 - 1 - 1 + -1 * I8 4.44/4.45 I14 <= I13 - 1 ==> I13 - 1 + -1 * I14 > I13 - 1 - 1 + -1 * I14 with I13 - 1 + -1 * I14 >= 0 4.44/4.45 4.44/4.45 We remove all the strictly oriented dependency pairs. 4.44/4.45 4.44/4.45 DP problem for innermost termination. 4.44/4.45 P = 4.44/4.45 f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 R = 4.44/4.45 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 4.44/4.45 f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 The dependency graph for this problem is: 4.44/4.45 3 -> 4.44/4.45 Where: 4.44/4.45 3) f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 4.44/4.45 We have the following SCCs. 4.44/4.45 4.44/7.42 EOF