/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) f3#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9] f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] f4#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] R = f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9] f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0] The dependency graph for this problem is: 0 -> 4 1 -> 2, 3 2 -> 1 3 -> 5 4 -> 5 5 -> 2, 3 Where: 0) f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 1) f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) 2) f3#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9] 3) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] 4) f4#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] We have the following SCCs. { 1, 2, 3, 5 } DP problem for innermost termination. P = f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) f3#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9] f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] R = f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9] f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = -1 * z3 + z5 + -1 * 0 NU[f3#(z1,z2,z3,z4,z5)] = -1 * z3 + z5 + -1 * 0 NU[f5#(z1,z2,z3,z4,z5)] = -1 * z3 + z5 + -1 * 0 This gives the following inequalities: ==> -1 * I2 + I4 + -1 * 0 >= -1 * I2 + I4 + -1 * 0 0 <= -1 * I7 + I9 ==> -1 * I7 + I9 + -1 * 0 > -1 * I7 + (-1 + I9) + -1 * 0 with -1 * I7 + I9 + -1 * 0 >= 0 1 - I12 + I14 <= 0 ==> -1 * I12 + I14 + -1 * 0 >= -1 * I12 + I14 + -1 * 0 0 <= -1 * I21 + I23 ==> -1 * I22 + I24 + -1 * 0 >= -1 * I22 + I24 + -1 * 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] R = f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9] f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0] The dependency graph for this problem is: 1 -> 3 3 -> 5 5 -> 3 Where: 1) f5#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) 3) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] 5) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] We have the following SCCs. { 3, 5 } DP problem for innermost termination. P = f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] R = f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9] f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = -1 * z2 + z4 + -1 * 0 NU[f3#(z1,z2,z3,z4,z5)] = -1 * z2 + (-1 + z4) + -1 * 0 This gives the following inequalities: 1 - I12 + I14 <= 0 ==> -1 * I11 + (-1 + I13) + -1 * 0 >= -1 * I11 + (-1 + I13) + -1 * 0 0 <= -1 * I21 + I23 ==> -1 * I21 + I23 + -1 * 0 > -1 * I21 + (-1 + I23) + -1 * 0 with -1 * I21 + I23 + -1 * 0 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] R = f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) f3(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, -1 + I9) [0 <= -1 * I7 + I9] f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] f4(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [0 <= -1 * I21 + I23] f1(I25, I26, I27, I28, I29) -> f2(rnd1, I26, I27, I28, I29) [rnd1 = rnd1 /\ 1 - I26 + I28 <= 0] The dependency graph for this problem is: 3 -> Where: 3) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I13, I14) [1 - I12 + I14 <= 0] We have the following SCCs.