/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE 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#(0, 1, rnd3, I3, I4) [rnd3 = rnd3] f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f6#(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f1#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f4#(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3#(I35, I36, I37, I38, I39) -> f1#(I35, I36, I37, I38, I39) f1#(I40, I41, I42, I43, I44) -> f3#(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(0, 1, rnd3, I3, I4) [rnd3 = rnd3] f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f4(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3(I35, I36, I37, I38, I39) -> f1(I35, I36, I37, I38, I39) f1(I40, I41, I42, I43, I44) -> f3(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] f1(I45, I46, I47, I48, I49) -> f2(I45, I46, I47, I48, I49) [I47 <= I49 /\ I48 <= I46 /\ 1 <= I45] The dependency graph for this problem is: 0 -> 1 1 -> 3, 5, 7, 9 2 -> 3, 5, 7, 9 3 -> 2 4 -> 3, 5, 7, 9 5 -> 4 6 -> 3, 5, 7, 9 7 -> 6 8 -> 3, 5, 7, 9 9 -> 8 Where: 0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 1) f7#(I0, I1, I2, I3, I4) -> f1#(0, 1, rnd3, I3, I4) [rnd3 = rnd3] 2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f1#(I10, I11, I12, I13, I14) -> f6#(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] 4) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5) f1#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] 6) f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 7) f1#(I30, I31, I32, I33, I34) -> f4#(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] 8) f3#(I35, I36, I37, I38, I39) -> f1#(I35, I36, I37, I38, I39) 9) f1#(I40, I41, I42, I43, I44) -> f3#(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8, 9 } 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, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f1#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f4#(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3#(I35, I36, I37, I38, I39) -> f1#(I35, I36, I37, I38, I39) f1#(I40, I41, I42, I43, I44) -> f3#(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(0, 1, rnd3, I3, I4) [rnd3 = rnd3] f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f4(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3(I35, I36, I37, I38, I39) -> f1(I35, I36, I37, I38, I39) f1(I40, I41, I42, I43, I44) -> f3(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] f1(I45, I46, I47, I48, I49) -> f2(I45, I46, I47, I48, I49) [I47 <= I49 /\ I48 <= I46 /\ 1 <= I45] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 NU[f4#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 NU[f5#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 NU[f1#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 NU[f6#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 This gives the following inequalities: ==> 0 + -1 * I5 >= 0 + -1 * I5 101 <= I12 /\ 1 <= I11 ==> 0 + -1 * I10 >= 0 + -1 * I10 ==> 0 + -1 * I15 >= 0 + -1 * I15 I22 <= 100 /\ 1 <= I21 ==> 0 + -1 * I20 >= 0 + -1 * I20 ==> 0 + -1 * I25 >= 0 + -1 * I25 101 <= I32 /\ 1 <= I31 /\ I30 <= 0 ==> 0 + -1 * I30 > 0 + -1 * 1 with 0 + -1 * I30 >= 0 ==> 0 + -1 * I35 >= 0 + -1 * I35 I42 <= 100 /\ 1 <= I41 /\ I40 <= 0 ==> 0 + -1 * I40 > 0 + -1 * 1 with 0 + -1 * I40 >= 0 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) f1#(I10, I11, I12, I13, I14) -> f6#(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f1#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f3#(I35, I36, I37, I38, I39) -> f1#(I35, I36, I37, I38, I39) R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(0, 1, rnd3, I3, I4) [rnd3 = rnd3] f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f4(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3(I35, I36, I37, I38, I39) -> f1(I35, I36, I37, I38, I39) f1(I40, I41, I42, I43, I44) -> f3(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] f1(I45, I46, I47, I48, I49) -> f2(I45, I46, I47, I48, I49) [I47 <= I49 /\ I48 <= I46 /\ 1 <= I45] The dependency graph for this problem is: 2 -> 3, 5 3 -> 2 4 -> 3, 5 5 -> 4 6 -> 3, 5 8 -> 3, 5 Where: 2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f1#(I10, I11, I12, I13, I14) -> f6#(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] 4) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5) f1#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] 6) f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 8) f3#(I35, I36, I37, I38, I39) -> f1#(I35, I36, I37, I38, I39) We have the following SCCs. { 2, 3, 4, 5 } 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, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f1#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(0, 1, rnd3, I3, I4) [rnd3 = rnd3] f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f4(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3(I35, I36, I37, I38, I39) -> f1(I35, I36, I37, I38, I39) f1(I40, I41, I42, I43, I44) -> f3(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] f1(I45, I46, I47, I48, I49) -> f2(I45, I46, I47, I48, I49) [I47 <= I49 /\ I48 <= I46 /\ 1 <= I45]