/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 = f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) f9#(I0, I1, I2, I3, I4) -> f7#(I0, I1, I2, I3, I4) f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) f7#(I10, I11, I12, I13, I14) -> f8#(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7#(I15, I16, I17, I18, I19) -> f5#(I15, I15, I17, I18, I19) [I19 <= I15] f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) f5#(I25, I26, I27, I28, I29) -> f6#(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I31, I33, I34) [I31 <= 0] f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) f3#(I40, I41, I42, I43, I44) -> f4#(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3#(I45, I46, I47, I48, I49) -> f1#(I45, I46, I47, I47, I49) [I49 <= I47] f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) f1#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, -1 + I58, I59) [1 <= I58] R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] The dependency graph for this problem is: 0 -> 1 1 -> 3, 4 2 -> 3, 4 3 -> 2 4 -> 6, 7 5 -> 6, 7 6 -> 5 7 -> 9, 10 8 -> 9, 10 9 -> 8 10 -> 12 11 -> 12 12 -> 11 Where: 0) f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 1) f9#(I0, I1, I2, I3, I4) -> f7#(I0, I1, I2, I3, I4) 2) f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) 3) f7#(I10, I11, I12, I13, I14) -> f8#(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 4) f7#(I15, I16, I17, I18, I19) -> f5#(I15, I15, I17, I18, I19) [I19 <= I15] 5) f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) 6) f5#(I25, I26, I27, I28, I29) -> f6#(I25, -1 + I26, I27, I28, I29) [1 <= I26] 7) f5#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I31, I33, I34) [I31 <= 0] 8) f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) 9) f3#(I40, I41, I42, I43, I44) -> f4#(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 10) f3#(I45, I46, I47, I48, I49) -> f1#(I45, I46, I47, I47, I49) [I49 <= I47] 11) f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) 12) f1#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, -1 + I58, I59) [1 <= I58] We have the following SCCs. { 2, 3 } { 5, 6 } { 8, 9 } { 11, 12 } DP problem for innermost termination. P = f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) f1#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, -1 + I58, I59) [1 <= I58] R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] We use the basic value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = z4 NU[f2#(z1,z2,z3,z4,z5)] = z4 This gives the following inequalities: ==> I53 (>! \union =) I53 1 <= I58 ==> I58 >! -1 + I58 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] The dependency graph for this problem is: 11 -> Where: 11) f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) We have the following SCCs. DP problem for innermost termination. P = f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) f3#(I40, I41, I42, I43, I44) -> f4#(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5)] = z5 + -1 * (1 + z3) NU[f4#(z1,z2,z3,z4,z5)] = z5 + -1 * (1 + z3) This gives the following inequalities: ==> I39 + -1 * (1 + I37) >= I39 + -1 * (1 + I37) 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 = f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] The dependency graph for this problem is: 8 -> Where: 8) f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) We have the following SCCs. DP problem for innermost termination. P = f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) f5#(I25, I26, I27, I28, I29) -> f6#(I25, -1 + I26, I27, I28, I29) [1 <= I26] R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] We use the basic value criterion with the projection function NU: NU[f5#(z1,z2,z3,z4,z5)] = z2 NU[f6#(z1,z2,z3,z4,z5)] = z2 This gives the following inequalities: ==> I21 (>! \union =) I21 1 <= I26 ==> I26 >! -1 + I26 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] The dependency graph for this problem is: 5 -> Where: 5) f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) We have the following SCCs. DP problem for innermost termination. P = f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) f7#(I10, I11, I12, I13, I14) -> f8#(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] We use the reverse value criterion with the projection function NU: NU[f7#(z1,z2,z3,z4,z5)] = z5 + -1 * (1 + z1) NU[f8#(z1,z2,z3,z4,z5)] = z5 + -1 * (1 + z1) This gives the following inequalities: ==> I9 + -1 * (1 + I5) >= I9 + -1 * (1 + I5) 1 + I10 <= I14 ==> I14 + -1 * (1 + I10) > I14 + -1 * (1 + (1 + I10)) with I14 + -1 * (1 + I10) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] The dependency graph for this problem is: 2 -> Where: 2) f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) We have the following SCCs.