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