/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 = f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) f10#(I0, I1, I2, I3, I4) -> f6#(I0, I1, I2, I3, rnd5) [rnd5 = rnd5] f2#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [0 <= I5] f2#(I10, I11, I12, I13, I14) -> f7#(I10, I11, I12, I13, I14) [1 + I10 <= 0] f9#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, I18, I19) [1 + I15 <= 1023] f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) [1023 <= I20] f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) f6#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) [1 + I39 <= 0] f6#(I40, I41, I42, I43, I44) -> f5#(I40, I41, I42, I43, I44) [1 <= I44] f6#(I45, I46, I47, I48, I49) -> f4#(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0] f5#(I50, I51, I52, I53, I54) -> f4#(I50, I51, I52, 0, I54) f4#(I55, I56, I57, I58, I59) -> f3#(I55, 0, I57, I58, I59) f1#(I60, I61, I62, I63, I64) -> f3#(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I68 <= I66] R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f6(I0, I1, I2, I3, rnd5) [rnd5 = rnd5] f2(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [0 <= I5] f2(I10, I11, I12, I13, I14) -> f7(I10, I11, I12, I13, I14) [1 + I10 <= 0] f9(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, I18, I19) [1 + I15 <= 1023] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) [1023 <= I20] f7(I25, I26, I27, I28, I29) -> f8(I25, I26, I27, I28, I29) f3(I30, I31, I32, I33, I34) -> f1(I30, I31, I32, I33, I34) f6(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) [1 + I39 <= 0] f6(I40, I41, I42, I43, I44) -> f5(I40, I41, I42, I43, I44) [1 <= I44] f6(I45, I46, I47, I48, I49) -> f4(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0] f5(I50, I51, I52, I53, I54) -> f4(I50, I51, I52, 0, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, 0, I57, I58, I59) f1(I60, I61, I62, I63, I64) -> f3(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I68 <= I66] The dependency graph for this problem is: 0 -> 1 1 -> 7, 8, 9 2 -> 4, 5 3 -> 4 -> 5 -> 6 -> 12, 13 7 -> 10 8 -> 10 9 -> 11 10 -> 11 11 -> 6 12 -> 6 13 -> 2, 3 Where: 0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 1) f10#(I0, I1, I2, I3, I4) -> f6#(I0, I1, I2, I3, rnd5) [rnd5 = rnd5] 2) f2#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [0 <= I5] 3) f2#(I10, I11, I12, I13, I14) -> f7#(I10, I11, I12, I13, I14) [1 + I10 <= 0] 4) f9#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, I18, I19) [1 + I15 <= 1023] 5) f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) [1023 <= I20] 6) f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) 7) f6#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) [1 + I39 <= 0] 8) f6#(I40, I41, I42, I43, I44) -> f5#(I40, I41, I42, I43, I44) [1 <= I44] 9) f6#(I45, I46, I47, I48, I49) -> f4#(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0] 10) f5#(I50, I51, I52, I53, I54) -> f4#(I50, I51, I52, 0, I54) 11) f4#(I55, I56, I57, I58, I59) -> f3#(I55, 0, I57, I58, I59) 12) f1#(I60, I61, I62, I63, I64) -> f3#(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] 13) f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I68 <= I66] We have the following SCCs. { 6, 12 } DP problem for innermost termination. P = f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) f1#(I60, I61, I62, I63, I64) -> f3#(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f6(I0, I1, I2, I3, rnd5) [rnd5 = rnd5] f2(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [0 <= I5] f2(I10, I11, I12, I13, I14) -> f7(I10, I11, I12, I13, I14) [1 + I10 <= 0] f9(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, I18, I19) [1 + I15 <= 1023] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) [1023 <= I20] f7(I25, I26, I27, I28, I29) -> f8(I25, I26, I27, I28, I29) f3(I30, I31, I32, I33, I34) -> f1(I30, I31, I32, I33, I34) f6(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) [1 + I39 <= 0] f6(I40, I41, I42, I43, I44) -> f5(I40, I41, I42, I43, I44) [1 <= I44] f6(I45, I46, I47, I48, I49) -> f4(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0] f5(I50, I51, I52, I53, I54) -> f4(I50, I51, I52, 0, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, 0, I57, I58, I59) f1(I60, I61, I62, I63, I64) -> f3(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I68 <= I66] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = z4 + -1 * z2 NU[f3#(z1,z2,z3,z4,z5)] = z4 + -1 * z2 This gives the following inequalities: ==> I33 + -1 * I31 >= I33 + -1 * I31 I61 <= I63 ==> I63 + -1 * I61 > I63 + -1 * (1 + I61) with I63 + -1 * I61 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f6(I0, I1, I2, I3, rnd5) [rnd5 = rnd5] f2(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [0 <= I5] f2(I10, I11, I12, I13, I14) -> f7(I10, I11, I12, I13, I14) [1 + I10 <= 0] f9(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, I18, I19) [1 + I15 <= 1023] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) [1023 <= I20] f7(I25, I26, I27, I28, I29) -> f8(I25, I26, I27, I28, I29) f3(I30, I31, I32, I33, I34) -> f1(I30, I31, I32, I33, I34) f6(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) [1 + I39 <= 0] f6(I40, I41, I42, I43, I44) -> f5(I40, I41, I42, I43, I44) [1 <= I44] f6(I45, I46, I47, I48, I49) -> f4(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0] f5(I50, I51, I52, I53, I54) -> f4(I50, I51, I52, 0, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, 0, I57, I58, I59) f1(I60, I61, I62, I63, I64) -> f3(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I68 <= I66] The dependency graph for this problem is: 6 -> Where: 6) f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) We have the following SCCs.