/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 = f7#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6#(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5#(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4#(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6#(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1#(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6#(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5#(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] f5#(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4#(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] f4#(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3#(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f4#(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1#(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] f3#(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4#(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) R = f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f2(I36, I37, I38, I39, I40, I41, I42, I43, I44) f6(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] f4(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f4(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] f3(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) f1(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f2(I100, I101, I102, I103, I104, I105, I106, I107, I108) [I108 = I105 /\ I107 = I104 /\ I106 = I103 /\ I105 = I105 /\ I104 = I104 /\ I103 = I103] The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5 1 -> 6 2 -> 7, 8 3 -> 9 4 -> 5 -> 6 6 -> 7, 8 7 -> 9 8 -> 9 -> 7, 8 Where: 0) f7#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6#(x1, x2, x3, x4, x5, x6, x7, x8, x9) 1) f6#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5#(I0, I1, I2, I3, I4, I5, I6, I7, I8) 2) f6#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 3) f6#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18, I19, I20, I21, I22, I23, I24, I25, I26) 4) f6#(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1#(I27, I28, I29, I30, I31, I32, I33, I34, I35) 5) f6#(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5#(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] 6) f5#(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4#(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] 7) f4#(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3#(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] 8) f4#(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1#(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] 9) f3#(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4#(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) We have the following SCCs. { 7, 9 } DP problem for innermost termination. P = f4#(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3#(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f3#(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4#(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) R = f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f2(I36, I37, I38, I39, I40, I41, I42, I43, I44) f6(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] f4(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f4(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] f3(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) f1(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f2(I100, I101, I102, I103, I104, I105, I106, I107, I108) [I108 = I105 /\ I107 = I104 /\ I106 = I103 /\ I105 = I105 /\ I104 = I104 /\ I103 = I103] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z7 + -1 * (1 + (1 + z8)) NU[f4#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z7 + -1 * (1 + z8) This gives the following inequalities: 1 + I74 <= I73 ==> I73 + -1 * (1 + I74) > I73 + -1 * (1 + (1 + I74)) with I73 + -1 * (1 + I74) >= 0 ==> I91 + -1 * (1 + (1 + I92)) >= I91 + -1 * (1 + (1 + I92)) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4#(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) R = f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f2(I36, I37, I38, I39, I40, I41, I42, I43, I44) f6(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] f4(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f4(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] f3(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) f1(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f2(I100, I101, I102, I103, I104, I105, I106, I107, I108) [I108 = I105 /\ I107 = I104 /\ I106 = I103 /\ I105 = I105 /\ I104 = I104 /\ I103 = I103] The dependency graph for this problem is: 9 -> Where: 9) f3#(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4#(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) We have the following SCCs.