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