0.00/0.49 YES 0.00/0.49 0.00/0.49 DP problem for innermost termination. 0.00/0.49 P = 0.00/0.49 init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 0.00/0.49 f4#(I0, I1, I2) -> f4#(I3, I4, I2) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 1 <= I1 /\ I4 + 3 <= I0 /\ I3 <= I1 /\ 1 <= I2 - 1 /\ I3 + 2 <= I0] 0.00/0.49 f3#(I5, I6, I7) -> f4#(I8, I9, I7) [0 <= y1 - 1 /\ 1 <= I7 - 1 /\ I8 <= I6 /\ I9 + 1 <= I6 /\ 0 <= I5 - 1 /\ 0 <= I6 - 1 /\ 0 <= I8 - 1 /\ -1 <= I9 - 1] 0.00/0.49 f2#(I10, I11, I12) -> f2#(I10 - 1, I11, I12 + 1) [I12 <= I11 - 1 /\ I10 - 1 <= I10 - 1 /\ 0 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I10 - 1] 0.00/0.49 f1#(I13, I14, I15) -> f3#(I16, I17, I18) [-1 <= I19 - 1 /\ 0 <= I14 - 1 /\ I16 <= I13 /\ I17 - 1 <= I13 /\ 0 <= I13 - 1 /\ 0 <= I16 - 1 /\ 1 <= I17 - 1] 0.00/0.49 f1#(I20, I21, I22) -> f3#(I23, I24, I25) [-1 <= I26 - 1 /\ 0 <= I21 - 1 /\ I23 <= I20 /\ 0 <= I20 - 1 /\ 0 <= I23 - 1 /\ 2 <= I24 - 1] 0.00/0.49 f1#(I27, I28, I29) -> f2#(I30, I28, 1) [0 <= I27 - 1 /\ 0 <= I28 - 1 /\ -1 <= I30 - 1] 0.00/0.49 R = 0.00/0.49 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 0.00/0.49 f4(I0, I1, I2) -> f4(I3, I4, I2) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 1 <= I1 /\ I4 + 3 <= I0 /\ I3 <= I1 /\ 1 <= I2 - 1 /\ I3 + 2 <= I0] 0.00/0.49 f3(I5, I6, I7) -> f4(I8, I9, I7) [0 <= y1 - 1 /\ 1 <= I7 - 1 /\ I8 <= I6 /\ I9 + 1 <= I6 /\ 0 <= I5 - 1 /\ 0 <= I6 - 1 /\ 0 <= I8 - 1 /\ -1 <= I9 - 1] 0.00/0.49 f2(I10, I11, I12) -> f2(I10 - 1, I11, I12 + 1) [I12 <= I11 - 1 /\ I10 - 1 <= I10 - 1 /\ 0 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I10 - 1] 0.00/0.49 f1(I13, I14, I15) -> f3(I16, I17, I18) [-1 <= I19 - 1 /\ 0 <= I14 - 1 /\ I16 <= I13 /\ I17 - 1 <= I13 /\ 0 <= I13 - 1 /\ 0 <= I16 - 1 /\ 1 <= I17 - 1] 0.00/0.49 f1(I20, I21, I22) -> f3(I23, I24, I25) [-1 <= I26 - 1 /\ 0 <= I21 - 1 /\ I23 <= I20 /\ 0 <= I20 - 1 /\ 0 <= I23 - 1 /\ 2 <= I24 - 1] 0.00/0.49 f1(I27, I28, I29) -> f2(I30, I28, 1) [0 <= I27 - 1 /\ 0 <= I28 - 1 /\ -1 <= I30 - 1] 0.00/0.49 0.00/0.49 The dependency graph for this problem is: 0.00/0.49 0 -> 4, 5, 6 0.00/0.49 1 -> 1 0.00/0.49 2 -> 1 0.00/0.49 3 -> 3 0.00/0.49 4 -> 2 0.00/0.49 5 -> 2 0.00/0.49 6 -> 3 0.00/0.49 Where: 0.00/0.49 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 0.00/0.49 1) f4#(I0, I1, I2) -> f4#(I3, I4, I2) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 1 <= I1 /\ I4 + 3 <= I0 /\ I3 <= I1 /\ 1 <= I2 - 1 /\ I3 + 2 <= I0] 0.00/0.49 2) f3#(I5, I6, I7) -> f4#(I8, I9, I7) [0 <= y1 - 1 /\ 1 <= I7 - 1 /\ I8 <= I6 /\ I9 + 1 <= I6 /\ 0 <= I5 - 1 /\ 0 <= I6 - 1 /\ 0 <= I8 - 1 /\ -1 <= I9 - 1] 0.00/0.49 3) f2#(I10, I11, I12) -> f2#(I10 - 1, I11, I12 + 1) [I12 <= I11 - 1 /\ I10 - 1 <= I10 - 1 /\ 0 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I10 - 1] 0.00/0.49 4) f1#(I13, I14, I15) -> f3#(I16, I17, I18) [-1 <= I19 - 1 /\ 0 <= I14 - 1 /\ I16 <= I13 /\ I17 - 1 <= I13 /\ 0 <= I13 - 1 /\ 0 <= I16 - 1 /\ 1 <= I17 - 1] 0.00/0.49 5) f1#(I20, I21, I22) -> f3#(I23, I24, I25) [-1 <= I26 - 1 /\ 0 <= I21 - 1 /\ I23 <= I20 /\ 0 <= I20 - 1 /\ 0 <= I23 - 1 /\ 2 <= I24 - 1] 0.00/0.49 6) f1#(I27, I28, I29) -> f2#(I30, I28, 1) [0 <= I27 - 1 /\ 0 <= I28 - 1 /\ -1 <= I30 - 1] 0.00/0.49 0.00/0.49 We have the following SCCs. 0.00/0.49 { 1 } 0.00/0.49 { 3 } 0.00/0.49 0.00/0.49 DP problem for innermost termination. 0.00/0.49 P = 0.00/0.49 f2#(I10, I11, I12) -> f2#(I10 - 1, I11, I12 + 1) [I12 <= I11 - 1 /\ I10 - 1 <= I10 - 1 /\ 0 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I10 - 1] 0.00/0.49 R = 0.00/0.49 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 0.00/0.49 f4(I0, I1, I2) -> f4(I3, I4, I2) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 1 <= I1 /\ I4 + 3 <= I0 /\ I3 <= I1 /\ 1 <= I2 - 1 /\ I3 + 2 <= I0] 0.00/0.49 f3(I5, I6, I7) -> f4(I8, I9, I7) [0 <= y1 - 1 /\ 1 <= I7 - 1 /\ I8 <= I6 /\ I9 + 1 <= I6 /\ 0 <= I5 - 1 /\ 0 <= I6 - 1 /\ 0 <= I8 - 1 /\ -1 <= I9 - 1] 0.00/0.49 f2(I10, I11, I12) -> f2(I10 - 1, I11, I12 + 1) [I12 <= I11 - 1 /\ I10 - 1 <= I10 - 1 /\ 0 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I10 - 1] 0.00/0.49 f1(I13, I14, I15) -> f3(I16, I17, I18) [-1 <= I19 - 1 /\ 0 <= I14 - 1 /\ I16 <= I13 /\ I17 - 1 <= I13 /\ 0 <= I13 - 1 /\ 0 <= I16 - 1 /\ 1 <= I17 - 1] 0.00/0.49 f1(I20, I21, I22) -> f3(I23, I24, I25) [-1 <= I26 - 1 /\ 0 <= I21 - 1 /\ I23 <= I20 /\ 0 <= I20 - 1 /\ 0 <= I23 - 1 /\ 2 <= I24 - 1] 0.00/0.49 f1(I27, I28, I29) -> f2(I30, I28, 1) [0 <= I27 - 1 /\ 0 <= I28 - 1 /\ -1 <= I30 - 1] 0.00/0.49 0.00/0.49 We use the basic value criterion with the projection function NU: 0.00/0.49 NU[f2#(z1,z2,z3)] = z1 0.00/0.49 0.00/0.49 This gives the following inequalities: 0.00/0.49 I12 <= I11 - 1 /\ I10 - 1 <= I10 - 1 /\ 0 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I10 - 1 ==> I10 >! I10 - 1 0.00/0.49 0.00/0.49 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/0.49 0.00/0.49 DP problem for innermost termination. 0.00/0.49 P = 0.00/0.49 f4#(I0, I1, I2) -> f4#(I3, I4, I2) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 1 <= I1 /\ I4 + 3 <= I0 /\ I3 <= I1 /\ 1 <= I2 - 1 /\ I3 + 2 <= I0] 0.00/0.49 R = 0.00/0.49 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 0.00/0.49 f4(I0, I1, I2) -> f4(I3, I4, I2) [-1 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 1 <= I1 /\ I4 + 3 <= I0 /\ I3 <= I1 /\ 1 <= I2 - 1 /\ I3 + 2 <= I0] 0.00/0.49 f3(I5, I6, I7) -> f4(I8, I9, I7) [0 <= y1 - 1 /\ 1 <= I7 - 1 /\ I8 <= I6 /\ I9 + 1 <= I6 /\ 0 <= I5 - 1 /\ 0 <= I6 - 1 /\ 0 <= I8 - 1 /\ -1 <= I9 - 1] 0.00/0.49 f2(I10, I11, I12) -> f2(I10 - 1, I11, I12 + 1) [I12 <= I11 - 1 /\ I10 - 1 <= I10 - 1 /\ 0 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I10 - 1] 0.00/0.49 f1(I13, I14, I15) -> f3(I16, I17, I18) [-1 <= I19 - 1 /\ 0 <= I14 - 1 /\ I16 <= I13 /\ I17 - 1 <= I13 /\ 0 <= I13 - 1 /\ 0 <= I16 - 1 /\ 1 <= I17 - 1] 0.00/0.49 f1(I20, I21, I22) -> f3(I23, I24, I25) [-1 <= I26 - 1 /\ 0 <= I21 - 1 /\ I23 <= I20 /\ 0 <= I20 - 1 /\ 0 <= I23 - 1 /\ 2 <= I24 - 1] 0.00/0.49 f1(I27, I28, I29) -> f2(I30, I28, 1) [0 <= I27 - 1 /\ 0 <= I28 - 1 /\ -1 <= I30 - 1] 0.00/0.49 0.00/0.49 We use the basic value criterion with the projection function NU: 0.00/0.49 NU[f4#(z1,z2,z3)] = z2 0.00/0.49 0.00/0.49 This gives the following inequalities: 0.00/0.49 -1 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 + 1 <= I1 /\ I4 + 3 <= I0 /\ I3 <= I1 /\ 1 <= I2 - 1 /\ I3 + 2 <= I0 ==> I1 >! I4 0.00/0.49 0.00/0.49 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.48 EOF