/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13] f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27] f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27] f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33] f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 3, 7 2 -> 3, 7 3 -> 2 4 -> 3, 7 5 -> 4 6 -> 5 7 -> 6 Where: 0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 1) f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f1#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13] 4) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 6) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27] 7) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33] We have the following SCCs. { 2, 3, 4, 5, 6, 7 } DP problem for innermost termination. P = f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13] f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27] f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27] f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33] f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1,x2,x3,x4)] = -x4 - 2 NU[f4#(x0,x1,x2,x3,x4)] = -x4 - 2 NU[f5#(x0,x1,x2,x3,x4)] = -x4 - 1 NU[f1#(x0,x1,x2,x3,x4)] = -x4 - 1 NU[f6#(x0,x1,x2,x3,x4)] = -x4 - 1 This gives the following inequalities: ==> -I9 - 1 >= -I9 - 1 -1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13 ==> -I14 - 1 > -(1 + I14) - 1 with -I14 - 1 >= 0 ==> -I19 - 1 >= -I19 - 1 ==> -I24 - 2 >= -(1 + I24) - 1 I27 = I27 ==> -I29 - 2 >= -I29 - 2 0 <= -1 - I33 ==> -I34 - 1 >= -I34 - 2 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27] f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27] f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33] f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0] The dependency graph for this problem is: 2 -> 7 4 -> 7 5 -> 4 6 -> 5 7 -> 6 Where: 2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 4) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 6) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27] 7) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33] We have the following SCCs. { 4, 5, 6, 7 } DP problem for innermost termination. P = f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27] f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I32, I33, I34) [0 <= -1 - I33] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27] f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33] f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1,x2,x3,x4)] = -x3 - 2 NU[f4#(x0,x1,x2,x3,x4)] = -x3 - 2 NU[f1#(x0,x1,x2,x3,x4)] = -x3 - 1 NU[f5#(x0,x1,x2,x3,x4)] = -x3 - 1 This gives the following inequalities: ==> -I18 - 1 >= -I18 - 1 ==> -I23 - 2 >= -(1 + I23) - 1 I27 = I27 ==> -I28 - 2 >= -I28 - 2 0 <= -1 - I33 ==> -I33 - 1 > -I33 - 2 with -I33 - 1 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, -1 * I11 + I13, 1 + I14) [-1 <= I14 /\ I14 <= -1 /\ 0 <= -1 - I13] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f4(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, 1 + I23, 1 + I24) f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I27 = I27] f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I32, I33, I34) [0 <= -1 - I33] f1(I35, I36, I37, I38, I39) -> f2(rnd1, I36, I37, I38, I39) [rnd1 = rnd1 /\ -1 * I38 <= 0] The dependency graph for this problem is: 4 -> 5 -> 4 6 -> 5 Where: 4) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5) f4#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, 1 + I23, 1 + I24) 6) f3#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I27 = I27] We have the following SCCs.