/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, I49, I50, I51, I52, rnd9) [rnd9 = rnd4 /\ rnd4 = rnd4] f5#(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4#(I60, I61, I62, I57, I58, I59, I60, I61, I60) f4#(I63, I64, I65, I66, I67, I68, I69, I70, I71) -> f3#(I69, I70, I71, I66, I67, I68, I69, I70, I71) [1 <= I70 /\ I70 <= I71] f4#(I72, I73, I74, I75, I76, I77, I78, I79, I80) -> f1#(I78, I79, I80, I75, I76, I77, I78, I79, I80) [1 + I80 <= I79] f4#(I81, I82, I83, I84, I85, I86, I87, I88, I89) -> f1#(I87, I88, I89, I84, I85, I86, I87, I88, I89) [I88 <= 0] f3#(I90, I91, I92, I93, I94, I95, I96, I97, I98) -> f4#(I96, I97, I98, I93, I94, I95, I96, I97, -1 * I97 + I98) 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, I49, I50, I51, I52, rnd9) [rnd9 = rnd4 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4(I60, I61, I62, I57, I58, I59, I60, I61, I60) f4(I63, I64, I65, I66, I67, I68, I69, I70, I71) -> f3(I69, I70, I71, I66, I67, I68, I69, I70, I71) [1 <= I70 /\ I70 <= I71] f4(I72, I73, I74, I75, I76, I77, I78, I79, I80) -> f1(I78, I79, I80, I75, I76, I77, I78, I79, I80) [1 + I80 <= I79] f4(I81, I82, I83, I84, I85, I86, I87, I88, I89) -> f1(I87, I88, I89, I84, I85, I86, I87, I88, I89) [I88 <= 0] f3(I90, I91, I92, I93, I94, I95, I96, I97, I98) -> f4(I96, I97, I98, I93, I94, I95, I96, I97, -1 * I97 + I98) f1(I99, I100, I101, I102, I103, I104, I105, I106, I107) -> f2(I105, I106, I107, I108, rnd5, rnd6, rnd7, rnd8, I109) [I109 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = I108 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ I108 = I108] The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5 1 -> 6 2 -> 7, 8, 9 3 -> 10 4 -> 5 -> 6 6 -> 7, 8, 9 7 -> 10 8 -> 9 -> 10 -> 7, 8, 9 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, I49, I50, I51, I52, rnd9) [rnd9 = rnd4 /\ rnd4 = rnd4] 6) f5#(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4#(I60, I61, I62, I57, I58, I59, I60, I61, I60) 7) f4#(I63, I64, I65, I66, I67, I68, I69, I70, I71) -> f3#(I69, I70, I71, I66, I67, I68, I69, I70, I71) [1 <= I70 /\ I70 <= I71] 8) f4#(I72, I73, I74, I75, I76, I77, I78, I79, I80) -> f1#(I78, I79, I80, I75, I76, I77, I78, I79, I80) [1 + I80 <= I79] 9) f4#(I81, I82, I83, I84, I85, I86, I87, I88, I89) -> f1#(I87, I88, I89, I84, I85, I86, I87, I88, I89) [I88 <= 0] 10) f3#(I90, I91, I92, I93, I94, I95, I96, I97, I98) -> f4#(I96, I97, I98, I93, I94, I95, I96, I97, -1 * I97 + I98) We have the following SCCs. { 7, 10 } DP problem for innermost termination. P = f4#(I63, I64, I65, I66, I67, I68, I69, I70, I71) -> f3#(I69, I70, I71, I66, I67, I68, I69, I70, I71) [1 <= I70 /\ I70 <= I71] f3#(I90, I91, I92, I93, I94, I95, I96, I97, I98) -> f4#(I96, I97, I98, I93, I94, I95, I96, I97, -1 * I97 + I98) 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, I49, I50, I51, I52, rnd9) [rnd9 = rnd4 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4(I60, I61, I62, I57, I58, I59, I60, I61, I60) f4(I63, I64, I65, I66, I67, I68, I69, I70, I71) -> f3(I69, I70, I71, I66, I67, I68, I69, I70, I71) [1 <= I70 /\ I70 <= I71] f4(I72, I73, I74, I75, I76, I77, I78, I79, I80) -> f1(I78, I79, I80, I75, I76, I77, I78, I79, I80) [1 + I80 <= I79] f4(I81, I82, I83, I84, I85, I86, I87, I88, I89) -> f1(I87, I88, I89, I84, I85, I86, I87, I88, I89) [I88 <= 0] f3(I90, I91, I92, I93, I94, I95, I96, I97, I98) -> f4(I96, I97, I98, I93, I94, I95, I96, I97, -1 * I97 + I98) f1(I99, I100, I101, I102, I103, I104, I105, I106, I107) -> f2(I105, I106, I107, I108, rnd5, rnd6, rnd7, rnd8, I109) [I109 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = I108 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ I108 = I108] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = -1 * z8 + z9 + -1 * z8 NU[f4#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z9 + -1 * z8 This gives the following inequalities: 1 <= I70 /\ I70 <= I71 ==> I71 + -1 * I70 > -1 * I70 + I71 + -1 * I70 with I71 + -1 * I70 >= 0 ==> -1 * I97 + I98 + -1 * I97 >= -1 * I97 + I98 + -1 * I97 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I90, I91, I92, I93, I94, I95, I96, I97, I98) -> f4#(I96, I97, I98, I93, I94, I95, I96, I97, -1 * I97 + I98) 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, I49, I50, I51, I52, rnd9) [rnd9 = rnd4 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4(I60, I61, I62, I57, I58, I59, I60, I61, I60) f4(I63, I64, I65, I66, I67, I68, I69, I70, I71) -> f3(I69, I70, I71, I66, I67, I68, I69, I70, I71) [1 <= I70 /\ I70 <= I71] f4(I72, I73, I74, I75, I76, I77, I78, I79, I80) -> f1(I78, I79, I80, I75, I76, I77, I78, I79, I80) [1 + I80 <= I79] f4(I81, I82, I83, I84, I85, I86, I87, I88, I89) -> f1(I87, I88, I89, I84, I85, I86, I87, I88, I89) [I88 <= 0] f3(I90, I91, I92, I93, I94, I95, I96, I97, I98) -> f4(I96, I97, I98, I93, I94, I95, I96, I97, -1 * I97 + I98) f1(I99, I100, I101, I102, I103, I104, I105, I106, I107) -> f2(I105, I106, I107, I108, rnd5, rnd6, rnd7, rnd8, I109) [I109 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = I108 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ I108 = I108] The dependency graph for this problem is: 10 -> Where: 10) f3#(I90, I91, I92, I93, I94, I95, I96, I97, I98) -> f4#(I96, I97, I98, I93, I94, I95, I96, I97, -1 * I97 + I98) We have the following SCCs.