/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 = f13#(x1, x2, x3, x4, x5, x6, x7) -> f12#(x1, x2, x3, x4, x5, x6, x7) f12#(I0, I1, I2, I3, I4, I5, I6) -> f6#(I0, I1, 0, I3, I4, I5, I6) f2#(I7, I8, I9, I10, I11, I12, I13) -> f11#(I7, I8, I9, I10, I11, I9, I13) f11#(I14, I15, I16, I17, I18, I19, I20) -> f10#(I14, I15, I16, I17, I18, I19, I20) f10#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, 1 + I23, I24, I25, I26, I27) f4#(I28, I29, I30, I31, I32, I33, I34) -> f8#(I28, I29, I30, I31, I32, I33, I34) f8#(I35, I36, I37, I38, I39, I40, I41) -> f7#(I35, I36, I37, I38, I39, I40, I38) [1 + I38 <= I36] f7#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, I50, I51, I52, I53, I54, I55) f6#(I56, I57, I58, I59, I60, I61, I62) -> f3#(I56, I57, I58, I59, I60, I61, I62) f5#(I63, I64, I65, I66, I67, I68, I69) -> f4#(I63, I64, I65, 1 + I66, I67, I68, I69) f3#(I70, I71, I72, I73, I74, I75, I76) -> f1#(I70, I71, I72, I73, I72, I75, I76) [1 + I72 <= I70] f3#(I77, I78, I79, I80, I81, I82, I83) -> f4#(I77, I78, I79, I77, I81, I82, I83) [I77 <= I79] f1#(I84, I85, I86, I87, I88, I89, I90) -> f2#(I84, I85, I86, I87, I88, I89, I90) R = f13(x1, x2, x3, x4, x5, x6, x7) -> f12(x1, x2, x3, x4, x5, x6, x7) f12(I0, I1, I2, I3, I4, I5, I6) -> f6(I0, I1, 0, I3, I4, I5, I6) f2(I7, I8, I9, I10, I11, I12, I13) -> f11(I7, I8, I9, I10, I11, I9, I13) f11(I14, I15, I16, I17, I18, I19, I20) -> f10(I14, I15, I16, I17, I18, I19, I20) f10(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, 1 + I23, I24, I25, I26, I27) f4(I28, I29, I30, I31, I32, I33, I34) -> f8(I28, I29, I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39, I40, I41) -> f7(I35, I36, I37, I38, I39, I40, I38) [1 + I38 <= I36] f8(I42, I43, I44, I45, I46, I47, I48) -> f9(I42, I43, I44, I45, I46, I47, I48) [I43 <= I45] f7(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, I50, I51, I52, I53, I54, I55) f6(I56, I57, I58, I59, I60, I61, I62) -> f3(I56, I57, I58, I59, I60, I61, I62) f5(I63, I64, I65, I66, I67, I68, I69) -> f4(I63, I64, I65, 1 + I66, I67, I68, I69) f3(I70, I71, I72, I73, I74, I75, I76) -> f1(I70, I71, I72, I73, I72, I75, I76) [1 + I72 <= I70] f3(I77, I78, I79, I80, I81, I82, I83) -> f4(I77, I78, I79, I77, I81, I82, I83) [I77 <= I79] f1(I84, I85, I86, I87, I88, I89, I90) -> f2(I84, I85, I86, I87, I88, I89, I90) The dependency graph for this problem is: 0 -> 1 1 -> 8 2 -> 3 3 -> 4 4 -> 8 5 -> 6 6 -> 7 7 -> 9 8 -> 10, 11 9 -> 5 10 -> 12 11 -> 5 12 -> 2 Where: 0) f13#(x1, x2, x3, x4, x5, x6, x7) -> f12#(x1, x2, x3, x4, x5, x6, x7) 1) f12#(I0, I1, I2, I3, I4, I5, I6) -> f6#(I0, I1, 0, I3, I4, I5, I6) 2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f11#(I7, I8, I9, I10, I11, I9, I13) 3) f11#(I14, I15, I16, I17, I18, I19, I20) -> f10#(I14, I15, I16, I17, I18, I19, I20) 4) f10#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, 1 + I23, I24, I25, I26, I27) 5) f4#(I28, I29, I30, I31, I32, I33, I34) -> f8#(I28, I29, I30, I31, I32, I33, I34) 6) f8#(I35, I36, I37, I38, I39, I40, I41) -> f7#(I35, I36, I37, I38, I39, I40, I38) [1 + I38 <= I36] 7) f7#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, I50, I51, I52, I53, I54, I55) 8) f6#(I56, I57, I58, I59, I60, I61, I62) -> f3#(I56, I57, I58, I59, I60, I61, I62) 9) f5#(I63, I64, I65, I66, I67, I68, I69) -> f4#(I63, I64, I65, 1 + I66, I67, I68, I69) 10) f3#(I70, I71, I72, I73, I74, I75, I76) -> f1#(I70, I71, I72, I73, I72, I75, I76) [1 + I72 <= I70] 11) f3#(I77, I78, I79, I80, I81, I82, I83) -> f4#(I77, I78, I79, I77, I81, I82, I83) [I77 <= I79] 12) f1#(I84, I85, I86, I87, I88, I89, I90) -> f2#(I84, I85, I86, I87, I88, I89, I90) We have the following SCCs. { 2, 3, 4, 8, 10, 12 } { 5, 6, 7, 9 } DP problem for innermost termination. P = f4#(I28, I29, I30, I31, I32, I33, I34) -> f8#(I28, I29, I30, I31, I32, I33, I34) f8#(I35, I36, I37, I38, I39, I40, I41) -> f7#(I35, I36, I37, I38, I39, I40, I38) [1 + I38 <= I36] f7#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, I50, I51, I52, I53, I54, I55) f5#(I63, I64, I65, I66, I67, I68, I69) -> f4#(I63, I64, I65, 1 + I66, I67, I68, I69) R = f13(x1, x2, x3, x4, x5, x6, x7) -> f12(x1, x2, x3, x4, x5, x6, x7) f12(I0, I1, I2, I3, I4, I5, I6) -> f6(I0, I1, 0, I3, I4, I5, I6) f2(I7, I8, I9, I10, I11, I12, I13) -> f11(I7, I8, I9, I10, I11, I9, I13) f11(I14, I15, I16, I17, I18, I19, I20) -> f10(I14, I15, I16, I17, I18, I19, I20) f10(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, 1 + I23, I24, I25, I26, I27) f4(I28, I29, I30, I31, I32, I33, I34) -> f8(I28, I29, I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39, I40, I41) -> f7(I35, I36, I37, I38, I39, I40, I38) [1 + I38 <= I36] f8(I42, I43, I44, I45, I46, I47, I48) -> f9(I42, I43, I44, I45, I46, I47, I48) [I43 <= I45] f7(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, I50, I51, I52, I53, I54, I55) f6(I56, I57, I58, I59, I60, I61, I62) -> f3(I56, I57, I58, I59, I60, I61, I62) f5(I63, I64, I65, I66, I67, I68, I69) -> f4(I63, I64, I65, 1 + I66, I67, I68, I69) f3(I70, I71, I72, I73, I74, I75, I76) -> f1(I70, I71, I72, I73, I72, I75, I76) [1 + I72 <= I70] f3(I77, I78, I79, I80, I81, I82, I83) -> f4(I77, I78, I79, I77, I81, I82, I83) [I77 <= I79] f1(I84, I85, I86, I87, I88, I89, I90) -> f2(I84, I85, I86, I87, I88, I89, I90) We use the extended value criterion with the projection function NU: NU[f5#(x0,x1,x2,x3,x4,x5,x6)] = x1 - x3 - 2 NU[f7#(x0,x1,x2,x3,x4,x5,x6)] = x1 - x3 - 2 NU[f8#(x0,x1,x2,x3,x4,x5,x6)] = x1 - x3 - 1 NU[f4#(x0,x1,x2,x3,x4,x5,x6)] = x1 - x3 - 1 This gives the following inequalities: ==> I29 - I31 - 1 >= I29 - I31 - 1 1 + I38 <= I36 ==> I36 - I38 - 1 > I36 - I38 - 2 with I36 - I38 - 1 >= 0 ==> I50 - I52 - 2 >= I50 - I52 - 2 ==> I64 - I66 - 2 >= I64 - (1 + I66) - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I28, I29, I30, I31, I32, I33, I34) -> f8#(I28, I29, I30, I31, I32, I33, I34) f7#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, I50, I51, I52, I53, I54, I55) f5#(I63, I64, I65, I66, I67, I68, I69) -> f4#(I63, I64, I65, 1 + I66, I67, I68, I69) R = f13(x1, x2, x3, x4, x5, x6, x7) -> f12(x1, x2, x3, x4, x5, x6, x7) f12(I0, I1, I2, I3, I4, I5, I6) -> f6(I0, I1, 0, I3, I4, I5, I6) f2(I7, I8, I9, I10, I11, I12, I13) -> f11(I7, I8, I9, I10, I11, I9, I13) f11(I14, I15, I16, I17, I18, I19, I20) -> f10(I14, I15, I16, I17, I18, I19, I20) f10(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, 1 + I23, I24, I25, I26, I27) f4(I28, I29, I30, I31, I32, I33, I34) -> f8(I28, I29, I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39, I40, I41) -> f7(I35, I36, I37, I38, I39, I40, I38) [1 + I38 <= I36] f8(I42, I43, I44, I45, I46, I47, I48) -> f9(I42, I43, I44, I45, I46, I47, I48) [I43 <= I45] f7(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, I50, I51, I52, I53, I54, I55) f6(I56, I57, I58, I59, I60, I61, I62) -> f3(I56, I57, I58, I59, I60, I61, I62) f5(I63, I64, I65, I66, I67, I68, I69) -> f4(I63, I64, I65, 1 + I66, I67, I68, I69) f3(I70, I71, I72, I73, I74, I75, I76) -> f1(I70, I71, I72, I73, I72, I75, I76) [1 + I72 <= I70] f3(I77, I78, I79, I80, I81, I82, I83) -> f4(I77, I78, I79, I77, I81, I82, I83) [I77 <= I79] f1(I84, I85, I86, I87, I88, I89, I90) -> f2(I84, I85, I86, I87, I88, I89, I90) The dependency graph for this problem is: 5 -> 7 -> 9 9 -> 5 Where: 5) f4#(I28, I29, I30, I31, I32, I33, I34) -> f8#(I28, I29, I30, I31, I32, I33, I34) 7) f7#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, I50, I51, I52, I53, I54, I55) 9) f5#(I63, I64, I65, I66, I67, I68, I69) -> f4#(I63, I64, I65, 1 + I66, I67, I68, I69) We have the following SCCs. DP problem for innermost termination. P = f2#(I7, I8, I9, I10, I11, I12, I13) -> f11#(I7, I8, I9, I10, I11, I9, I13) f11#(I14, I15, I16, I17, I18, I19, I20) -> f10#(I14, I15, I16, I17, I18, I19, I20) f10#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, 1 + I23, I24, I25, I26, I27) f6#(I56, I57, I58, I59, I60, I61, I62) -> f3#(I56, I57, I58, I59, I60, I61, I62) f3#(I70, I71, I72, I73, I74, I75, I76) -> f1#(I70, I71, I72, I73, I72, I75, I76) [1 + I72 <= I70] f1#(I84, I85, I86, I87, I88, I89, I90) -> f2#(I84, I85, I86, I87, I88, I89, I90) R = f13(x1, x2, x3, x4, x5, x6, x7) -> f12(x1, x2, x3, x4, x5, x6, x7) f12(I0, I1, I2, I3, I4, I5, I6) -> f6(I0, I1, 0, I3, I4, I5, I6) f2(I7, I8, I9, I10, I11, I12, I13) -> f11(I7, I8, I9, I10, I11, I9, I13) f11(I14, I15, I16, I17, I18, I19, I20) -> f10(I14, I15, I16, I17, I18, I19, I20) f10(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, 1 + I23, I24, I25, I26, I27) f4(I28, I29, I30, I31, I32, I33, I34) -> f8(I28, I29, I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39, I40, I41) -> f7(I35, I36, I37, I38, I39, I40, I38) [1 + I38 <= I36] f8(I42, I43, I44, I45, I46, I47, I48) -> f9(I42, I43, I44, I45, I46, I47, I48) [I43 <= I45] f7(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, I50, I51, I52, I53, I54, I55) f6(I56, I57, I58, I59, I60, I61, I62) -> f3(I56, I57, I58, I59, I60, I61, I62) f5(I63, I64, I65, I66, I67, I68, I69) -> f4(I63, I64, I65, 1 + I66, I67, I68, I69) f3(I70, I71, I72, I73, I74, I75, I76) -> f1(I70, I71, I72, I73, I72, I75, I76) [1 + I72 <= I70] f3(I77, I78, I79, I80, I81, I82, I83) -> f4(I77, I78, I79, I77, I81, I82, I83) [I77 <= I79] f1(I84, I85, I86, I87, I88, I89, I90) -> f2(I84, I85, I86, I87, I88, I89, I90) We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3,x4,x5,x6)] = x0 - x2 - 2 NU[f3#(x0,x1,x2,x3,x4,x5,x6)] = x0 - x2 - 1 NU[f6#(x0,x1,x2,x3,x4,x5,x6)] = x0 - x2 - 1 NU[f10#(x0,x1,x2,x3,x4,x5,x6)] = x0 - x2 - 2 NU[f11#(x0,x1,x2,x3,x4,x5,x6)] = x0 - x2 - 2 NU[f2#(x0,x1,x2,x3,x4,x5,x6)] = x0 - x2 - 2 This gives the following inequalities: ==> I7 - I9 - 2 >= I7 - I9 - 2 ==> I14 - I16 - 2 >= I14 - I16 - 2 ==> I21 - I23 - 2 >= I21 - (1 + I23) - 1 ==> I56 - I58 - 1 >= I56 - I58 - 1 1 + I72 <= I70 ==> I70 - I72 - 1 > I70 - I72 - 2 with I70 - I72 - 1 >= 0 ==> I84 - I86 - 2 >= I84 - I86 - 2 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I7, I8, I9, I10, I11, I12, I13) -> f11#(I7, I8, I9, I10, I11, I9, I13) f11#(I14, I15, I16, I17, I18, I19, I20) -> f10#(I14, I15, I16, I17, I18, I19, I20) f10#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, 1 + I23, I24, I25, I26, I27) f6#(I56, I57, I58, I59, I60, I61, I62) -> f3#(I56, I57, I58, I59, I60, I61, I62) f1#(I84, I85, I86, I87, I88, I89, I90) -> f2#(I84, I85, I86, I87, I88, I89, I90) R = f13(x1, x2, x3, x4, x5, x6, x7) -> f12(x1, x2, x3, x4, x5, x6, x7) f12(I0, I1, I2, I3, I4, I5, I6) -> f6(I0, I1, 0, I3, I4, I5, I6) f2(I7, I8, I9, I10, I11, I12, I13) -> f11(I7, I8, I9, I10, I11, I9, I13) f11(I14, I15, I16, I17, I18, I19, I20) -> f10(I14, I15, I16, I17, I18, I19, I20) f10(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, 1 + I23, I24, I25, I26, I27) f4(I28, I29, I30, I31, I32, I33, I34) -> f8(I28, I29, I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39, I40, I41) -> f7(I35, I36, I37, I38, I39, I40, I38) [1 + I38 <= I36] f8(I42, I43, I44, I45, I46, I47, I48) -> f9(I42, I43, I44, I45, I46, I47, I48) [I43 <= I45] f7(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, I50, I51, I52, I53, I54, I55) f6(I56, I57, I58, I59, I60, I61, I62) -> f3(I56, I57, I58, I59, I60, I61, I62) f5(I63, I64, I65, I66, I67, I68, I69) -> f4(I63, I64, I65, 1 + I66, I67, I68, I69) f3(I70, I71, I72, I73, I74, I75, I76) -> f1(I70, I71, I72, I73, I72, I75, I76) [1 + I72 <= I70] f3(I77, I78, I79, I80, I81, I82, I83) -> f4(I77, I78, I79, I77, I81, I82, I83) [I77 <= I79] f1(I84, I85, I86, I87, I88, I89, I90) -> f2(I84, I85, I86, I87, I88, I89, I90) The dependency graph for this problem is: 2 -> 3 3 -> 4 4 -> 8 8 -> 12 -> 2 Where: 2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f11#(I7, I8, I9, I10, I11, I9, I13) 3) f11#(I14, I15, I16, I17, I18, I19, I20) -> f10#(I14, I15, I16, I17, I18, I19, I20) 4) f10#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, 1 + I23, I24, I25, I26, I27) 8) f6#(I56, I57, I58, I59, I60, I61, I62) -> f3#(I56, I57, I58, I59, I60, I61, I62) 12) f1#(I84, I85, I86, I87, I88, I89, I90) -> f2#(I84, I85, I86, I87, I88, I89, I90) We have the following SCCs.