/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, x6) -> f1#(x1, x2, x3, x4, x5, x6) f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] f4#(I18, I19, I20, I21, I22, I23) -> f3#(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] f2#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] R = f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] The dependency graph for this problem is: 0 -> 6 1 -> 2, 3 2 -> 1 3 -> 5 4 -> 2, 3 5 -> 2 6 -> 5 Where: 0) f6#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 1) f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 2) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 3) f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] 4) f4#(I18, I19, I20, I21, I22, I23) -> f3#(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] 5) f2#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] 6) f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] We have the following SCCs. { 1, 2, 3, 5 } DP problem for innermost termination. P = f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] f2#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] R = f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] We use the extended value criterion with the projection function NU: NU[f2#(x0,x1,x2,x3,x4,x5)] = x0 - 1 NU[f3#(x0,x1,x2,x3,x4,x5)] = x0 - x3 - 1 NU[f5#(x0,x1,x2,x3,x4,x5)] = x0 - x3 - 1 This gives the following inequalities: ==> I0 - I3 - 1 >= I0 - I3 - 1 1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3 ==> I6 - I9 - 1 >= (-1 + I6) - (-1 + I9) - 1 I15 <= 0 /\ I15 <= 0 ==> I12 - I15 - 1 >= I12 - 1 1 <= I24 /\ 1 <= 5000 /\ 1 <= I24 ==> I24 - 1 > I24 - 5000 - 1 with I24 - 1 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] R = f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] The dependency graph for this problem is: 1 -> 2, 3 2 -> 1 3 -> Where: 1) f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 2) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 3) f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] R = f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] We use the basic value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6)] = z4 NU[f5#(z1,z2,z3,z4,z5,z6)] = z4 This gives the following inequalities: ==> I3 (>! \union =) I3 1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3 ==> I9 >! -1 + I9 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) R = f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] The dependency graph for this problem is: 1 -> Where: 1) f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) We have the following SCCs.