/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 = f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) f6#(I0, I1, I2, I3, I4, I5, I6) -> f1#(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] f2#(I14, I15, I16, I17, I18, I19, I20) -> f5#(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) f3#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) R = f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) The dependency graph for this problem is: 0 -> 1 1 -> 6 2 -> 6 3 -> 4 4 -> 5 5 -> 4 6 -> 2, 3 Where: 0) f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) 1) f6#(I0, I1, I2, I3, I4, I5, I6) -> f1#(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] 2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] 3) f2#(I14, I15, I16, I17, I18, I19, I20) -> f5#(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] 4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 5) f3#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] 6) f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) We have the following SCCs. { 2, 6 } { 4, 5 } DP problem for innermost termination. P = f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) f3#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] R = f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = z3 + -1 * (1 + z2) NU[f5#(z1,z2,z3,z4,z5,z6,z7)] = z3 + -1 * (1 + z2) This gives the following inequalities: ==> I23 + -1 * (1 + I22) >= I23 + -1 * (1 + I22) 1 + I29 <= I30 ==> I30 + -1 * (1 + I29) > I30 + -1 * (1 + (1 + I29)) with I30 + -1 * (1 + I29) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) R = f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) The dependency graph for this problem is: 4 -> Where: 4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) We have the following SCCs. DP problem for innermost termination. P = f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) R = f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z1) NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z1) This gives the following inequalities: 1 + I7 <= I10 ==> I10 + -1 * (1 + I7) > I10 + -1 * (1 + (1 + I7)) with I10 + -1 * (1 + I7) >= 0 ==> I45 + -1 * (1 + I42) >= I45 + -1 * (1 + I42) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) R = f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) The dependency graph for this problem is: 6 -> Where: 6) f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) We have the following SCCs.