/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 = f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) f8#(I0, I1, I2, I3, I4, I5) -> f7#(I0, I1, I2, I3, I4, I5) f8#(I6, I7, I8, I9, I10, I11) -> f6#(I6, I7, I8, I9, I10, I11) f8#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, I16, I17) f8#(I18, I19, I20, I21, I22, I23) -> f4#(I18, I19, I20, I21, I22, I23) f8#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, I28, I29) f8#(I30, I31, I32, I33, I34, I35) -> f1#(I30, I31, I32, I33, I34, I35) f8#(I42, I43, I44, I45, I46, I47) -> f7#(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f7#(I48, I49, I50, I51, I52, I53) -> f6#(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] f6#(I56, I57, I58, I59, I60, I61) -> f5#(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f6#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, I68, 5) [100 <= I68] f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] f4#(I78, I79, I80, I81, I82, I83) -> f3#(I82, I83, I80, I81, I82, I83) [I83 <= 20] f4#(I84, I85, I86, I87, I88, I89) -> f1#(I88, I89, I86, I87, I88, I89) [21 <= I89] f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68] f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20] f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89] f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102] The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5, 6, 7 1 -> 8 2 -> 9, 10 3 -> 11 4 -> 12, 13 5 -> 14 6 -> 7 -> 8 8 -> 9 9 -> 11 10 -> 12 11 -> 9, 10 12 -> 14 13 -> 14 -> 12, 13 Where: 0) f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) 1) f8#(I0, I1, I2, I3, I4, I5) -> f7#(I0, I1, I2, I3, I4, I5) 2) f8#(I6, I7, I8, I9, I10, I11) -> f6#(I6, I7, I8, I9, I10, I11) 3) f8#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, I16, I17) 4) f8#(I18, I19, I20, I21, I22, I23) -> f4#(I18, I19, I20, I21, I22, I23) 5) f8#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, I28, I29) 6) f8#(I30, I31, I32, I33, I34, I35) -> f1#(I30, I31, I32, I33, I34, I35) 7) f8#(I42, I43, I44, I45, I46, I47) -> f7#(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] 8) f7#(I48, I49, I50, I51, I52, I53) -> f6#(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] 9) f6#(I56, I57, I58, I59, I60, I61) -> f5#(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] 10) f6#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, I68, 5) [100 <= I68] 11) f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] 12) f4#(I78, I79, I80, I81, I82, I83) -> f3#(I82, I83, I80, I81, I82, I83) [I83 <= 20] 13) f4#(I84, I85, I86, I87, I88, I89) -> f1#(I88, I89, I86, I87, I88, I89) [21 <= I89] 14) f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) We have the following SCCs. { 9, 11 } { 12, 14 } DP problem for innermost termination. P = f4#(I78, I79, I80, I81, I82, I83) -> f3#(I82, I83, I80, I81, I82, I83) [I83 <= 20] f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68] f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20] f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89] f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6)] = 20 + -1 * (3 + z6) NU[f4#(z1,z2,z3,z4,z5,z6)] = 20 + -1 * z6 This gives the following inequalities: I83 <= 20 ==> 20 + -1 * I83 > 20 + -1 * (3 + I83) with 20 + -1 * I83 >= 0 ==> 20 + -1 * (3 + I95) >= 20 + -1 * (3 + I95) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68] f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20] f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89] f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102] The dependency graph for this problem is: 14 -> Where: 14) f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) We have the following SCCs. DP problem for innermost termination. P = f6#(I56, I57, I58, I59, I60, I61) -> f5#(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68] f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20] f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89] f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102] We use the reverse value criterion with the projection function NU: NU[f5#(z1,z2,z3,z4,z5,z6)] = 99 + -1 * (1 + z5) NU[f6#(z1,z2,z3,z4,z5,z6)] = 99 + -1 * z5 This gives the following inequalities: I63 = I62 /\ I60 <= 99 /\ I62 = I62 ==> 99 + -1 * I60 > 99 + -1 * (1 + I60) with 99 + -1 * I60 >= 0 I77 = I76 /\ I76 = I76 ==> 99 + -1 * (1 + I74) >= 99 + -1 * (1 + I74) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68] f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20] f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89] f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102] The dependency graph for this problem is: 11 -> Where: 11) f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] We have the following SCCs.