/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) f10#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f5#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [1 + I5 <= I7] f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10] f5#(I15, I16, I17, I18, I19) -> f8#(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] f9#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) f8#(I35, I36, I37, I38, I39) -> f7#(-1 + I35, 0, I37, I38, I39) f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) [1 <= I62] f1#(I70, I71, I72, I73, I74) -> f2#(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7] f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10] f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46] f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0] f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62] f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67] f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] The dependency graph for this problem is: 0 -> 1 1 -> 11 2 -> 5, 6 3 -> 5, 6 4 -> 8 5 -> 8 6 -> 10 7 -> 9 8 -> 7 9 -> 7 10 -> 2, 3, 4 11 -> 12 12 -> 10 Where: 0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 1) f10#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] 2) f5#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [1 + I5 <= I7] 3) f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10] 4) f5#(I15, I16, I17, I18, I19) -> f8#(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] 5) f9#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 6) f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) 7) f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 8) f8#(I35, I36, I37, I38, I39) -> f7#(-1 + I35, 0, I37, I38, I39) 9) f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] 10) f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) 11) f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) [1 <= I62] 12) f1#(I70, I71, I72, I73, I74) -> f2#(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] We have the following SCCs. { 2, 3, 6, 10 } { 7, 9 } DP problem for innermost termination. P = f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7] f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10] f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46] f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0] f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62] f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67] f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] We use the reverse value criterion with the projection function NU: NU[f6#(z1,z2,z3,z4,z5)] = z1 + -1 * (1 + z2) NU[f7#(z1,z2,z3,z4,z5)] = z1 + -1 * (1 + z2) This gives the following inequalities: ==> I30 + -1 * (1 + I31) >= I30 + -1 * (1 + I31) 1 + I41 <= I40 ==> I40 + -1 * (1 + I41) > I40 + -1 * (1 + (1 + I41)) with I40 + -1 * (1 + I41) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7] f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10] f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46] f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0] f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62] f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67] f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] The dependency graph for this problem is: 7 -> Where: 7) f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) We have the following SCCs. DP problem for innermost termination. P = f5#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [1 + I5 <= I7] f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10] f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7] f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10] f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46] f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0] f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62] f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67] f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] We use the extended value criterion with the projection function NU: NU[f2#(x0,x1,x2,x3,x4)] = -x0 + x2 - 1 NU[f9#(x0,x1,x2,x3,x4)] = -x0 + x2 - 2 NU[f5#(x0,x1,x2,x3,x4)] = -x0 + x2 - 1 This gives the following inequalities: 1 + I5 <= I7 ==> -I5 + I7 - 1 > -I5 + I7 - 2 with -I5 + I7 - 1 >= 0 1 + I12 <= I10 ==> -I10 + I12 - 1 >= -I10 + I12 - 2 ==> -I25 + I27 - 2 >= -(1 + I25) + I27 - 1 ==> -I50 + I52 - 1 >= -I50 + I52 - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10] f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7] f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10] f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46] f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0] f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62] f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67] f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100]