/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 = f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) f8#(I0, I1, I2, I3, I4, I5) -> f1#(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] f2#(I6, I7, I8, I9, I10, I11) -> f1#(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] f2#(I12, I13, I14, I15, I16, I17) -> f6#(0, I13, I14, I15, I16, I17) [I14 <= I12] f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) f7#(I24, I25, I26, I27, I28, I29) -> f6#(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] f7#(I30, I31, I32, I33, I34, I35) -> f5#(0, I31, I32, I33, I34, I35) [I32 <= I30] f6#(I36, I37, I38, I39, I40, I41) -> f7#(I36, I37, I38, I39, I40, I41) f3#(I42, I43, I44, I45, I46, I47) -> f5#(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] f1#(I54, I55, I56, I57, I58, I59) -> f2#(I54, I55, I56, I57, I58, I59) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) The dependency graph for this problem is: 0 -> 1 1 -> 9 2 -> 9 3 -> 7 4 -> 8 5 -> 7 6 -> 4 7 -> 5, 6 8 -> 4 9 -> 2, 3 Where: 0) f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) 1) f8#(I0, I1, I2, I3, I4, I5) -> f1#(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] 2) f2#(I6, I7, I8, I9, I10, I11) -> f1#(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] 3) f2#(I12, I13, I14, I15, I16, I17) -> f6#(0, I13, I14, I15, I16, I17) [I14 <= I12] 4) f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 5) f7#(I24, I25, I26, I27, I28, I29) -> f6#(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] 6) f7#(I30, I31, I32, I33, I34, I35) -> f5#(0, I31, I32, I33, I34, I35) [I32 <= I30] 7) f6#(I36, I37, I38, I39, I40, I41) -> f7#(I36, I37, I38, I39, I40, I41) 8) f3#(I42, I43, I44, I45, I46, I47) -> f5#(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] 9) f1#(I54, I55, I56, I57, I58, I59) -> f2#(I54, I55, I56, I57, I58, I59) We have the following SCCs. { 2, 9 } { 5, 7 } { 4, 8 } DP problem for innermost termination. P = f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) f3#(I42, I43, I44, I45, I46, I47) -> f5#(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z1) NU[f5#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z1) This gives the following inequalities: ==> I20 + -1 * (1 + I18) >= I20 + -1 * (1 + I18) 1 + I42 <= I44 ==> I44 + -1 * (1 + I42) > I44 + -1 * (1 + (1 + I42)) with I44 + -1 * (1 + I42) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) The dependency graph for this problem is: 4 -> Where: 4) f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) We have the following SCCs. DP problem for innermost termination. P = f7#(I24, I25, I26, I27, I28, I29) -> f6#(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] f6#(I36, I37, I38, I39, I40, I41) -> f7#(I36, I37, I38, I39, I40, I41) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) We use the reverse value criterion with the projection function NU: NU[f6#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z1) NU[f7#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z1) This gives the following inequalities: 1 + I24 <= I26 ==> I26 + -1 * (1 + I24) > I26 + -1 * (1 + (1 + I24)) with I26 + -1 * (1 + I24) >= 0 ==> I38 + -1 * (1 + I36) >= I38 + -1 * (1 + I36) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I36, I37, I38, I39, I40, I41) -> f7#(I36, I37, I38, I39, I40, I41) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) The dependency graph for this problem is: 7 -> Where: 7) f6#(I36, I37, I38, I39, I40, I41) -> f7#(I36, I37, I38, I39, I40, I41) We have the following SCCs. DP problem for innermost termination. P = f2#(I6, I7, I8, I9, I10, I11) -> f1#(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] f1#(I54, I55, I56, I57, I58, I59) -> f2#(I54, I55, I56, I57, I58, I59) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z1) NU[f2#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z1) This gives the following inequalities: 1 + I6 <= I8 ==> I8 + -1 * (1 + I6) > I8 + -1 * (1 + (1 + I6)) with I8 + -1 * (1 + I6) >= 0 ==> I56 + -1 * (1 + I54) >= I56 + -1 * (1 + I54) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I54, I55, I56, I57, I58, I59) -> f2#(I54, I55, I56, I57, I58, I59) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) The dependency graph for this problem is: 9 -> Where: 9) f1#(I54, I55, I56, I57, I58, I59) -> f2#(I54, I55, I56, I57, I58, I59) We have the following SCCs.