/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) f10#(I0, I1, I2, I3, I4) -> f4#(I0, 1, I2, I3, I4) f2#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I5, I9) f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) [0 <= I13] f6#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I15) [1 + I18 <= 0] f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) f7#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, -1 + I29) [0 <= I29] f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) f4#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0] f4#(I50, I51, I52, I53, I54) -> f2#(I50, I51, I52, I53, I54) [1 + I51 <= 0] f4#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, I58, I59) [1 <= I56] f1#(I60, I61, I62, I63, I64) -> f3#(I60, I61, -1 + I62, I63, I64) [0 <= I62] f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I67 <= 0] R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13] f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29] f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0] f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0] f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0] f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56] f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 11 2 -> 7 3 -> 7 4 -> 5 5 -> 6 6 -> 5 7 -> 3, 4 8 -> 12, 13 9 -> 8 10 -> 2 11 -> 2 12 -> 8 13 -> 2 Where: 0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 1) f10#(I0, I1, I2, I3, I4) -> f4#(I0, 1, I2, I3, I4) 2) f2#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I5, I9) 3) f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) [0 <= I13] 4) f6#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I15) [1 + I18 <= 0] 5) f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) 6) f7#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, -1 + I29) [0 <= I29] 7) f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) 8) f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) 9) f4#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0] 10) f4#(I50, I51, I52, I53, I54) -> f2#(I50, I51, I52, I53, I54) [1 + I51 <= 0] 11) f4#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, I58, I59) [1 <= I56] 12) f1#(I60, I61, I62, I63, I64) -> f3#(I60, I61, -1 + I62, I63, I64) [0 <= I62] 13) f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I67 <= 0] We have the following SCCs. { 8, 12 } { 3, 7 } { 5, 6 } DP problem for innermost termination. P = f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) f7#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, -1 + I29) [0 <= I29] R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13] f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29] f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0] f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0] f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0] f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56] f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0] We use the basic value criterion with the projection function NU: NU[f7#(z1,z2,z3,z4,z5)] = z5 NU[f9#(z1,z2,z3,z4,z5)] = z5 This gives the following inequalities: ==> I24 (>! \union =) I24 0 <= I29 ==> I29 >! -1 + I29 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13] f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29] f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0] f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0] f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0] f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56] f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0] The dependency graph for this problem is: 5 -> Where: 5) f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) We have the following SCCs. DP problem for innermost termination. P = f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) [0 <= I13] f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13] f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29] f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0] f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0] f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0] f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56] f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0] We use the basic value criterion with the projection function NU: NU[f5#(z1,z2,z3,z4,z5)] = z4 NU[f6#(z1,z2,z3,z4,z5)] = z4 This gives the following inequalities: 0 <= I13 ==> I13 >! -1 + I13 ==> I38 (>! \union =) I38 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13] f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29] f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0] f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0] f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0] f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56] f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0] The dependency graph for this problem is: 7 -> Where: 7) f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) We have the following SCCs. DP problem for innermost termination. P = f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) f1#(I60, I61, I62, I63, I64) -> f3#(I60, I61, -1 + I62, I63, I64) [0 <= I62] R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13] f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29] f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0] f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0] f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0] f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56] f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0] We use the basic value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = z3 NU[f3#(z1,z2,z3,z4,z5)] = z3 This gives the following inequalities: ==> I42 (>! \union =) I42 0 <= I62 ==> I62 >! -1 + I62 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13] f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29] f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0] f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0] f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0] f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56] f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0] The dependency graph for this problem is: 8 -> Where: 8) f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) We have the following SCCs.