/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 = f12#(x1, x2, x3, x4, x5, x6, x7) -> f11#(x1, x2, x3, x4, x5, x6, x7) f11#(I0, I1, I2, I3, I4, I5, I6) -> f4#(0, I1, I2, I3, I4, I5, I6) f3#(I7, I8, I9, I10, I11, I12, I13) -> f4#(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= 10] f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [10 <= I14] f2#(I21, I22, I23, I24, I25, I26, I27) -> f8#(0, I22, I23, I24, I25, I26, I27) f10#(I28, I29, I30, I31, I32, I33, I34) -> f9#(I28, I29, rnd3, I31, rnd5, I33, I28) [rnd5 = rnd3 /\ rnd3 = rnd3 /\ 1 + I28 <= 10] f10#(I35, I36, I37, I38, I39, I40, I41) -> f5#(I35, I36, I37, I38, I39, I40, I41) [10 <= I35] f8#(I42, I43, I44, I45, I46, I47, I48) -> f10#(I42, I43, I44, I45, I46, I47, I48) f9#(I49, I50, I51, I52, I53, I54, I55) -> f7#(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] f9#(I56, I57, I58, I59, I60, I61, I62) -> f7#(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] f9#(I63, I64, I65, I66, I67, I68, I69) -> f5#(I63, I64, I65, I66, I67, I68, I69) [0 <= I67 /\ I67 <= 0] f7#(I70, I71, I72, I73, I74, I75, I76) -> f8#(1 + I70, I71, I72, I73, I74, I75, I76) f4#(I84, I85, I86, I87, I88, I89, I90) -> f1#(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] f1#(I91, I92, I93, I94, I95, I96, I97) -> f3#(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] f1#(I98, I99, I100, I101, I102, I103, I104) -> f3#(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] f1#(I105, I106, I107, I108, I109, I110, I111) -> f2#(I105, I106, I107, I108, I109, I110, I111) [0 <= I108 /\ I108 <= 0] R = f12(x1, x2, x3, x4, x5, x6, x7) -> f11(x1, x2, x3, x4, x5, x6, x7) f11(I0, I1, I2, I3, I4, I5, I6) -> f4(0, I1, I2, I3, I4, I5, I6) f3(I7, I8, I9, I10, I11, I12, I13) -> f4(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= 10] f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [10 <= I14] f2(I21, I22, I23, I24, I25, I26, I27) -> f8(0, I22, I23, I24, I25, I26, I27) f10(I28, I29, I30, I31, I32, I33, I34) -> f9(I28, I29, rnd3, I31, rnd5, I33, I28) [rnd5 = rnd3 /\ rnd3 = rnd3 /\ 1 + I28 <= 10] f10(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, I36, I37, I38, I39, I40, I41) [10 <= I35] f8(I42, I43, I44, I45, I46, I47, I48) -> f10(I42, I43, I44, I45, I46, I47, I48) f9(I49, I50, I51, I52, I53, I54, I55) -> f7(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] f9(I56, I57, I58, I59, I60, I61, I62) -> f7(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] f9(I63, I64, I65, I66, I67, I68, I69) -> f5(I63, I64, I65, I66, I67, I68, I69) [0 <= I67 /\ I67 <= 0] f7(I70, I71, I72, I73, I74, I75, I76) -> f8(1 + I70, I71, I72, I73, I74, I75, I76) f5(I77, I78, I79, I80, I81, I82, I83) -> f6(I77, I78, I79, I80, I81, I82, I83) f4(I84, I85, I86, I87, I88, I89, I90) -> f1(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] f1(I91, I92, I93, I94, I95, I96, I97) -> f3(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] f1(I98, I99, I100, I101, I102, I103, I104) -> f3(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] f1(I105, I106, I107, I108, I109, I110, I111) -> f2(I105, I106, I107, I108, I109, I110, I111) [0 <= I108 /\ I108 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 12 2 -> 12 3 -> 4 4 -> 7 5 -> 8, 9, 10 6 -> 7 -> 5, 6 8 -> 11 9 -> 11 10 -> 11 -> 7 12 -> 13, 14, 15 13 -> 2, 3 14 -> 2, 3 15 -> 4 Where: 0) f12#(x1, x2, x3, x4, x5, x6, x7) -> f11#(x1, x2, x3, x4, x5, x6, x7) 1) f11#(I0, I1, I2, I3, I4, I5, I6) -> f4#(0, I1, I2, I3, I4, I5, I6) 2) f3#(I7, I8, I9, I10, I11, I12, I13) -> f4#(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= 10] 3) f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [10 <= I14] 4) f2#(I21, I22, I23, I24, I25, I26, I27) -> f8#(0, I22, I23, I24, I25, I26, I27) 5) f10#(I28, I29, I30, I31, I32, I33, I34) -> f9#(I28, I29, rnd3, I31, rnd5, I33, I28) [rnd5 = rnd3 /\ rnd3 = rnd3 /\ 1 + I28 <= 10] 6) f10#(I35, I36, I37, I38, I39, I40, I41) -> f5#(I35, I36, I37, I38, I39, I40, I41) [10 <= I35] 7) f8#(I42, I43, I44, I45, I46, I47, I48) -> f10#(I42, I43, I44, I45, I46, I47, I48) 8) f9#(I49, I50, I51, I52, I53, I54, I55) -> f7#(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] 9) f9#(I56, I57, I58, I59, I60, I61, I62) -> f7#(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] 10) f9#(I63, I64, I65, I66, I67, I68, I69) -> f5#(I63, I64, I65, I66, I67, I68, I69) [0 <= I67 /\ I67 <= 0] 11) f7#(I70, I71, I72, I73, I74, I75, I76) -> f8#(1 + I70, I71, I72, I73, I74, I75, I76) 12) f4#(I84, I85, I86, I87, I88, I89, I90) -> f1#(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] 13) f1#(I91, I92, I93, I94, I95, I96, I97) -> f3#(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] 14) f1#(I98, I99, I100, I101, I102, I103, I104) -> f3#(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] 15) f1#(I105, I106, I107, I108, I109, I110, I111) -> f2#(I105, I106, I107, I108, I109, I110, I111) [0 <= I108 /\ I108 <= 0] We have the following SCCs. { 2, 12, 13, 14 } { 5, 7, 8, 9, 11 } DP problem for innermost termination. P = f10#(I28, I29, I30, I31, I32, I33, I34) -> f9#(I28, I29, rnd3, I31, rnd5, I33, I28) [rnd5 = rnd3 /\ rnd3 = rnd3 /\ 1 + I28 <= 10] f8#(I42, I43, I44, I45, I46, I47, I48) -> f10#(I42, I43, I44, I45, I46, I47, I48) f9#(I49, I50, I51, I52, I53, I54, I55) -> f7#(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] f9#(I56, I57, I58, I59, I60, I61, I62) -> f7#(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] f7#(I70, I71, I72, I73, I74, I75, I76) -> f8#(1 + I70, I71, I72, I73, I74, I75, I76) R = f12(x1, x2, x3, x4, x5, x6, x7) -> f11(x1, x2, x3, x4, x5, x6, x7) f11(I0, I1, I2, I3, I4, I5, I6) -> f4(0, I1, I2, I3, I4, I5, I6) f3(I7, I8, I9, I10, I11, I12, I13) -> f4(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= 10] f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [10 <= I14] f2(I21, I22, I23, I24, I25, I26, I27) -> f8(0, I22, I23, I24, I25, I26, I27) f10(I28, I29, I30, I31, I32, I33, I34) -> f9(I28, I29, rnd3, I31, rnd5, I33, I28) [rnd5 = rnd3 /\ rnd3 = rnd3 /\ 1 + I28 <= 10] f10(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, I36, I37, I38, I39, I40, I41) [10 <= I35] f8(I42, I43, I44, I45, I46, I47, I48) -> f10(I42, I43, I44, I45, I46, I47, I48) f9(I49, I50, I51, I52, I53, I54, I55) -> f7(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] f9(I56, I57, I58, I59, I60, I61, I62) -> f7(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] f9(I63, I64, I65, I66, I67, I68, I69) -> f5(I63, I64, I65, I66, I67, I68, I69) [0 <= I67 /\ I67 <= 0] f7(I70, I71, I72, I73, I74, I75, I76) -> f8(1 + I70, I71, I72, I73, I74, I75, I76) f5(I77, I78, I79, I80, I81, I82, I83) -> f6(I77, I78, I79, I80, I81, I82, I83) f4(I84, I85, I86, I87, I88, I89, I90) -> f1(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] f1(I91, I92, I93, I94, I95, I96, I97) -> f3(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] f1(I98, I99, I100, I101, I102, I103, I104) -> f3(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] f1(I105, I106, I107, I108, I109, I110, I111) -> f2(I105, I106, I107, I108, I109, I110, I111) [0 <= I108 /\ I108 <= 0] We use the extended value criterion with the projection function NU: NU[f7#(x0,x1,x2,x3,x4,x5,x6)] = -x0 + 8 NU[f8#(x0,x1,x2,x3,x4,x5,x6)] = -x0 + 9 NU[f9#(x0,x1,x2,x3,x4,x5,x6)] = -x0 + 8 NU[f10#(x0,x1,x2,x3,x4,x5,x6)] = -x0 + 9 This gives the following inequalities: rnd5 = rnd3 /\ rnd3 = rnd3 /\ 1 + I28 <= 10 ==> -I28 + 9 > -I28 + 8 with -I28 + 9 >= 0 ==> -I42 + 9 >= -I42 + 9 1 + I53 <= 0 ==> -I49 + 8 >= -I49 + 8 1 <= I60 ==> -I56 + 8 >= -I56 + 8 ==> -I70 + 8 >= -(1 + I70) + 9 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f8#(I42, I43, I44, I45, I46, I47, I48) -> f10#(I42, I43, I44, I45, I46, I47, I48) f9#(I49, I50, I51, I52, I53, I54, I55) -> f7#(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] f9#(I56, I57, I58, I59, I60, I61, I62) -> f7#(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] f7#(I70, I71, I72, I73, I74, I75, I76) -> f8#(1 + I70, I71, I72, I73, I74, I75, I76) R = f12(x1, x2, x3, x4, x5, x6, x7) -> f11(x1, x2, x3, x4, x5, x6, x7) f11(I0, I1, I2, I3, I4, I5, I6) -> f4(0, I1, I2, I3, I4, I5, I6) f3(I7, I8, I9, I10, I11, I12, I13) -> f4(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= 10] f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [10 <= I14] f2(I21, I22, I23, I24, I25, I26, I27) -> f8(0, I22, I23, I24, I25, I26, I27) f10(I28, I29, I30, I31, I32, I33, I34) -> f9(I28, I29, rnd3, I31, rnd5, I33, I28) [rnd5 = rnd3 /\ rnd3 = rnd3 /\ 1 + I28 <= 10] f10(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, I36, I37, I38, I39, I40, I41) [10 <= I35] f8(I42, I43, I44, I45, I46, I47, I48) -> f10(I42, I43, I44, I45, I46, I47, I48) f9(I49, I50, I51, I52, I53, I54, I55) -> f7(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] f9(I56, I57, I58, I59, I60, I61, I62) -> f7(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] f9(I63, I64, I65, I66, I67, I68, I69) -> f5(I63, I64, I65, I66, I67, I68, I69) [0 <= I67 /\ I67 <= 0] f7(I70, I71, I72, I73, I74, I75, I76) -> f8(1 + I70, I71, I72, I73, I74, I75, I76) f5(I77, I78, I79, I80, I81, I82, I83) -> f6(I77, I78, I79, I80, I81, I82, I83) f4(I84, I85, I86, I87, I88, I89, I90) -> f1(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] f1(I91, I92, I93, I94, I95, I96, I97) -> f3(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] f1(I98, I99, I100, I101, I102, I103, I104) -> f3(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] f1(I105, I106, I107, I108, I109, I110, I111) -> f2(I105, I106, I107, I108, I109, I110, I111) [0 <= I108 /\ I108 <= 0] The dependency graph for this problem is: 7 -> 8 -> 11 9 -> 11 11 -> 7 Where: 7) f8#(I42, I43, I44, I45, I46, I47, I48) -> f10#(I42, I43, I44, I45, I46, I47, I48) 8) f9#(I49, I50, I51, I52, I53, I54, I55) -> f7#(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] 9) f9#(I56, I57, I58, I59, I60, I61, I62) -> f7#(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] 11) f7#(I70, I71, I72, I73, I74, I75, I76) -> f8#(1 + I70, I71, I72, I73, I74, I75, I76) We have the following SCCs. DP problem for innermost termination. P = f3#(I7, I8, I9, I10, I11, I12, I13) -> f4#(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= 10] f4#(I84, I85, I86, I87, I88, I89, I90) -> f1#(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] f1#(I91, I92, I93, I94, I95, I96, I97) -> f3#(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] f1#(I98, I99, I100, I101, I102, I103, I104) -> f3#(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] R = f12(x1, x2, x3, x4, x5, x6, x7) -> f11(x1, x2, x3, x4, x5, x6, x7) f11(I0, I1, I2, I3, I4, I5, I6) -> f4(0, I1, I2, I3, I4, I5, I6) f3(I7, I8, I9, I10, I11, I12, I13) -> f4(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= 10] f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [10 <= I14] f2(I21, I22, I23, I24, I25, I26, I27) -> f8(0, I22, I23, I24, I25, I26, I27) f10(I28, I29, I30, I31, I32, I33, I34) -> f9(I28, I29, rnd3, I31, rnd5, I33, I28) [rnd5 = rnd3 /\ rnd3 = rnd3 /\ 1 + I28 <= 10] f10(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, I36, I37, I38, I39, I40, I41) [10 <= I35] f8(I42, I43, I44, I45, I46, I47, I48) -> f10(I42, I43, I44, I45, I46, I47, I48) f9(I49, I50, I51, I52, I53, I54, I55) -> f7(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] f9(I56, I57, I58, I59, I60, I61, I62) -> f7(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] f9(I63, I64, I65, I66, I67, I68, I69) -> f5(I63, I64, I65, I66, I67, I68, I69) [0 <= I67 /\ I67 <= 0] f7(I70, I71, I72, I73, I74, I75, I76) -> f8(1 + I70, I71, I72, I73, I74, I75, I76) f5(I77, I78, I79, I80, I81, I82, I83) -> f6(I77, I78, I79, I80, I81, I82, I83) f4(I84, I85, I86, I87, I88, I89, I90) -> f1(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] f1(I91, I92, I93, I94, I95, I96, I97) -> f3(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] f1(I98, I99, I100, I101, I102, I103, I104) -> f3(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] f1(I105, I106, I107, I108, I109, I110, I111) -> f2(I105, I106, I107, I108, I109, I110, I111) [0 <= I108 /\ I108 <= 0] We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3,x4,x5,x6)] = -x0 + 9 NU[f4#(x0,x1,x2,x3,x4,x5,x6)] = -x0 + 9 NU[f3#(x0,x1,x2,x3,x4,x5,x6)] = -x0 + 9 This gives the following inequalities: 1 + I7 <= 10 ==> -I7 + 9 > -(1 + I7) + 9 with -I7 + 9 >= 0 rnd4 = rnd2 /\ rnd2 = rnd2 ==> -I84 + 9 >= -I84 + 9 1 + I94 <= 0 ==> -I91 + 9 >= -I91 + 9 1 <= I101 ==> -I98 + 9 >= -I98 + 9 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I84, I85, I86, I87, I88, I89, I90) -> f1#(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] f1#(I91, I92, I93, I94, I95, I96, I97) -> f3#(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] f1#(I98, I99, I100, I101, I102, I103, I104) -> f3#(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] R = f12(x1, x2, x3, x4, x5, x6, x7) -> f11(x1, x2, x3, x4, x5, x6, x7) f11(I0, I1, I2, I3, I4, I5, I6) -> f4(0, I1, I2, I3, I4, I5, I6) f3(I7, I8, I9, I10, I11, I12, I13) -> f4(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= 10] f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [10 <= I14] f2(I21, I22, I23, I24, I25, I26, I27) -> f8(0, I22, I23, I24, I25, I26, I27) f10(I28, I29, I30, I31, I32, I33, I34) -> f9(I28, I29, rnd3, I31, rnd5, I33, I28) [rnd5 = rnd3 /\ rnd3 = rnd3 /\ 1 + I28 <= 10] f10(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, I36, I37, I38, I39, I40, I41) [10 <= I35] f8(I42, I43, I44, I45, I46, I47, I48) -> f10(I42, I43, I44, I45, I46, I47, I48) f9(I49, I50, I51, I52, I53, I54, I55) -> f7(I49, I50, I51, I52, I53, I54, I55) [1 + I53 <= 0] f9(I56, I57, I58, I59, I60, I61, I62) -> f7(I56, I57, I58, I59, I60, I61, I62) [1 <= I60] f9(I63, I64, I65, I66, I67, I68, I69) -> f5(I63, I64, I65, I66, I67, I68, I69) [0 <= I67 /\ I67 <= 0] f7(I70, I71, I72, I73, I74, I75, I76) -> f8(1 + I70, I71, I72, I73, I74, I75, I76) f5(I77, I78, I79, I80, I81, I82, I83) -> f6(I77, I78, I79, I80, I81, I82, I83) f4(I84, I85, I86, I87, I88, I89, I90) -> f1(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] f1(I91, I92, I93, I94, I95, I96, I97) -> f3(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] f1(I98, I99, I100, I101, I102, I103, I104) -> f3(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] f1(I105, I106, I107, I108, I109, I110, I111) -> f2(I105, I106, I107, I108, I109, I110, I111) [0 <= I108 /\ I108 <= 0] The dependency graph for this problem is: 12 -> 13, 14 13 -> 14 -> Where: 12) f4#(I84, I85, I86, I87, I88, I89, I90) -> f1#(I84, rnd2, I86, rnd4, I88, I84, I90) [rnd4 = rnd2 /\ rnd2 = rnd2] 13) f1#(I91, I92, I93, I94, I95, I96, I97) -> f3#(I91, I92, I93, I94, I95, I96, I97) [1 + I94 <= 0] 14) f1#(I98, I99, I100, I101, I102, I103, I104) -> f3#(I98, I99, I100, I101, I102, I103, I104) [1 <= I101] We have the following SCCs.