/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f10#(x1, x2, x3, x4, x5, x6) -> f9#(x1, x2, x3, x4, x5, x6) f9#(I0, I1, I2, I3, I4, I5) -> f4#(rnd1, 1, 0, rnd4, 1, I5) [rnd4 = 1 /\ rnd1 = 1] f5#(I6, I7, I8, I9, I10, I11) -> f8#(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] f5#(I12, I13, I14, I15, I16, I17) -> f4#(2 + I12, I13, I14, -10 + I15, I16, I17) [I12 <= I15] f8#(I18, I19, I20, I21, I22, I23) -> f7#(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] f8#(I25, I26, I27, I28, I29, I30) -> f7#(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] f7#(I31, I32, I33, I34, I35, I36) -> f6#(I31, I32, I33, I34, I35, I36) [10 <= I34] f7#(I37, I38, I39, I40, I41, I42) -> f3#(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] f6#(I43, I44, I45, I46, I47, I48) -> f3#(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] f6#(I49, I50, I51, I52, I53, I54) -> f3#(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] f3#(I55, I56, I57, I58, I59, I60) -> f5#(I55, I56, I57, I58, I59, I60) f4#(I61, I62, I63, I64, I65, I66) -> f1#(I61, I62, I63, I64, I65, I66) f1#(I67, I68, I69, I70, I71, I72) -> f3#(I67, I68, I69, I70, I71, I72) [1 + I67 <= 30] R = f10(x1, x2, x3, x4, x5, x6) -> f9(x1, x2, x3, x4, x5, x6) f9(I0, I1, I2, I3, I4, I5) -> f4(rnd1, 1, 0, rnd4, 1, I5) [rnd4 = 1 /\ rnd1 = 1] f5(I6, I7, I8, I9, I10, I11) -> f8(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] f5(I12, I13, I14, I15, I16, I17) -> f4(2 + I12, I13, I14, -10 + I15, I16, I17) [I12 <= I15] f8(I18, I19, I20, I21, I22, I23) -> f7(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] f8(I25, I26, I27, I28, I29, I30) -> f7(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] f7(I31, I32, I33, I34, I35, I36) -> f6(I31, I32, I33, I34, I35, I36) [10 <= I34] f7(I37, I38, I39, I40, I41, I42) -> f3(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] f6(I43, I44, I45, I46, I47, I48) -> f3(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] f6(I49, I50, I51, I52, I53, I54) -> f3(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] f3(I55, I56, I57, I58, I59, I60) -> f5(I55, I56, I57, I58, I59, I60) f4(I61, I62, I63, I64, I65, I66) -> f1(I61, I62, I63, I64, I65, I66) f1(I67, I68, I69, I70, I71, I72) -> f3(I67, I68, I69, I70, I71, I72) [1 + I67 <= 30] f1(I73, I74, I75, I76, I77, I78) -> f2(I73, I74, rnd3, I76, I77, 1) [rnd3 = 1 /\ 30 <= I73] The dependency graph for this problem is: 0 -> 1 1 -> 11 2 -> 4, 5 3 -> 11 4 -> 6, 7 5 -> 7 6 -> 8, 9 7 -> 10 8 -> 10 9 -> 10 10 -> 2, 3 11 -> 12 12 -> 10 Where: 0) f10#(x1, x2, x3, x4, x5, x6) -> f9#(x1, x2, x3, x4, x5, x6) 1) f9#(I0, I1, I2, I3, I4, I5) -> f4#(rnd1, 1, 0, rnd4, 1, I5) [rnd4 = 1 /\ rnd1 = 1] 2) f5#(I6, I7, I8, I9, I10, I11) -> f8#(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] 3) f5#(I12, I13, I14, I15, I16, I17) -> f4#(2 + I12, I13, I14, -10 + I15, I16, I17) [I12 <= I15] 4) f8#(I18, I19, I20, I21, I22, I23) -> f7#(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] 5) f8#(I25, I26, I27, I28, I29, I30) -> f7#(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] 6) f7#(I31, I32, I33, I34, I35, I36) -> f6#(I31, I32, I33, I34, I35, I36) [10 <= I34] 7) f7#(I37, I38, I39, I40, I41, I42) -> f3#(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] 8) f6#(I43, I44, I45, I46, I47, I48) -> f3#(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] 9) f6#(I49, I50, I51, I52, I53, I54) -> f3#(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] 10) f3#(I55, I56, I57, I58, I59, I60) -> f5#(I55, I56, I57, I58, I59, I60) 11) f4#(I61, I62, I63, I64, I65, I66) -> f1#(I61, I62, I63, I64, I65, I66) 12) f1#(I67, I68, I69, I70, I71, I72) -> f3#(I67, I68, I69, I70, I71, I72) [1 + I67 <= 30] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } DP problem for innermost termination. P = f5#(I6, I7, I8, I9, I10, I11) -> f8#(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] f5#(I12, I13, I14, I15, I16, I17) -> f4#(2 + I12, I13, I14, -10 + I15, I16, I17) [I12 <= I15] f8#(I18, I19, I20, I21, I22, I23) -> f7#(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] f8#(I25, I26, I27, I28, I29, I30) -> f7#(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] f7#(I31, I32, I33, I34, I35, I36) -> f6#(I31, I32, I33, I34, I35, I36) [10 <= I34] f7#(I37, I38, I39, I40, I41, I42) -> f3#(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] f6#(I43, I44, I45, I46, I47, I48) -> f3#(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] f6#(I49, I50, I51, I52, I53, I54) -> f3#(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] f3#(I55, I56, I57, I58, I59, I60) -> f5#(I55, I56, I57, I58, I59, I60) f4#(I61, I62, I63, I64, I65, I66) -> f1#(I61, I62, I63, I64, I65, I66) f1#(I67, I68, I69, I70, I71, I72) -> f3#(I67, I68, I69, I70, I71, I72) [1 + I67 <= 30] R = f10(x1, x2, x3, x4, x5, x6) -> f9(x1, x2, x3, x4, x5, x6) f9(I0, I1, I2, I3, I4, I5) -> f4(rnd1, 1, 0, rnd4, 1, I5) [rnd4 = 1 /\ rnd1 = 1] f5(I6, I7, I8, I9, I10, I11) -> f8(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] f5(I12, I13, I14, I15, I16, I17) -> f4(2 + I12, I13, I14, -10 + I15, I16, I17) [I12 <= I15] f8(I18, I19, I20, I21, I22, I23) -> f7(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] f8(I25, I26, I27, I28, I29, I30) -> f7(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] f7(I31, I32, I33, I34, I35, I36) -> f6(I31, I32, I33, I34, I35, I36) [10 <= I34] f7(I37, I38, I39, I40, I41, I42) -> f3(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] f6(I43, I44, I45, I46, I47, I48) -> f3(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] f6(I49, I50, I51, I52, I53, I54) -> f3(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] f3(I55, I56, I57, I58, I59, I60) -> f5(I55, I56, I57, I58, I59, I60) f4(I61, I62, I63, I64, I65, I66) -> f1(I61, I62, I63, I64, I65, I66) f1(I67, I68, I69, I70, I71, I72) -> f3(I67, I68, I69, I70, I71, I72) [1 + I67 <= 30] f1(I73, I74, I75, I76, I77, I78) -> f2(I73, I74, rnd3, I76, I77, 1) [rnd3 = 1 /\ 30 <= I73] We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3,x4,x5)] = -x0 + 29 NU[f3#(x0,x1,x2,x3,x4,x5)] = -x0 + 28 NU[f6#(x0,x1,x2,x3,x4,x5)] = -x0 + 27 NU[f7#(x0,x1,x2,x3,x4,x5)] = -x0 + 27 NU[f4#(x0,x1,x2,x3,x4,x5)] = -x0 + 29 NU[f8#(x0,x1,x2,x3,x4,x5)] = -x0 + 27 NU[f5#(x0,x1,x2,x3,x4,x5)] = -x0 + 28 This gives the following inequalities: 1 + I9 <= I6 ==> -I6 + 28 >= -I6 + 27 I12 <= I15 ==> -I12 + 28 >= -(2 + I12) + 29 I24 = I24 /\ 6 <= I21 ==> -I18 + 27 >= -I18 + 27 I28 <= 5 ==> -I25 + 27 >= -I25 + 27 10 <= I34 ==> -I31 + 27 >= -I31 + 27 1 + I40 <= 10 ==> -I37 + 27 >= -(1 + I37) + 28 I46 <= 12 ==> -I43 + 27 >= -(10 + I43) + 28 13 <= I52 ==> -I49 + 27 >= -(1 + I49) + 28 ==> -I55 + 28 >= -I55 + 28 ==> -I61 + 29 >= -I61 + 29 1 + I67 <= 30 ==> -I67 + 29 > -I67 + 28 with -I67 + 29 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I6, I7, I8, I9, I10, I11) -> f8#(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] f5#(I12, I13, I14, I15, I16, I17) -> f4#(2 + I12, I13, I14, -10 + I15, I16, I17) [I12 <= I15] f8#(I18, I19, I20, I21, I22, I23) -> f7#(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] f8#(I25, I26, I27, I28, I29, I30) -> f7#(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] f7#(I31, I32, I33, I34, I35, I36) -> f6#(I31, I32, I33, I34, I35, I36) [10 <= I34] f7#(I37, I38, I39, I40, I41, I42) -> f3#(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] f6#(I43, I44, I45, I46, I47, I48) -> f3#(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] f6#(I49, I50, I51, I52, I53, I54) -> f3#(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] f3#(I55, I56, I57, I58, I59, I60) -> f5#(I55, I56, I57, I58, I59, I60) f4#(I61, I62, I63, I64, I65, I66) -> f1#(I61, I62, I63, I64, I65, I66) R = f10(x1, x2, x3, x4, x5, x6) -> f9(x1, x2, x3, x4, x5, x6) f9(I0, I1, I2, I3, I4, I5) -> f4(rnd1, 1, 0, rnd4, 1, I5) [rnd4 = 1 /\ rnd1 = 1] f5(I6, I7, I8, I9, I10, I11) -> f8(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] f5(I12, I13, I14, I15, I16, I17) -> f4(2 + I12, I13, I14, -10 + I15, I16, I17) [I12 <= I15] f8(I18, I19, I20, I21, I22, I23) -> f7(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] f8(I25, I26, I27, I28, I29, I30) -> f7(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] f7(I31, I32, I33, I34, I35, I36) -> f6(I31, I32, I33, I34, I35, I36) [10 <= I34] f7(I37, I38, I39, I40, I41, I42) -> f3(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] f6(I43, I44, I45, I46, I47, I48) -> f3(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] f6(I49, I50, I51, I52, I53, I54) -> f3(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] f3(I55, I56, I57, I58, I59, I60) -> f5(I55, I56, I57, I58, I59, I60) f4(I61, I62, I63, I64, I65, I66) -> f1(I61, I62, I63, I64, I65, I66) f1(I67, I68, I69, I70, I71, I72) -> f3(I67, I68, I69, I70, I71, I72) [1 + I67 <= 30] f1(I73, I74, I75, I76, I77, I78) -> f2(I73, I74, rnd3, I76, I77, 1) [rnd3 = 1 /\ 30 <= I73] The dependency graph for this problem is: 2 -> 4, 5 3 -> 11 4 -> 6, 7 5 -> 7 6 -> 8, 9 7 -> 10 8 -> 10 9 -> 10 10 -> 2, 3 11 -> Where: 2) f5#(I6, I7, I8, I9, I10, I11) -> f8#(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] 3) f5#(I12, I13, I14, I15, I16, I17) -> f4#(2 + I12, I13, I14, -10 + I15, I16, I17) [I12 <= I15] 4) f8#(I18, I19, I20, I21, I22, I23) -> f7#(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] 5) f8#(I25, I26, I27, I28, I29, I30) -> f7#(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] 6) f7#(I31, I32, I33, I34, I35, I36) -> f6#(I31, I32, I33, I34, I35, I36) [10 <= I34] 7) f7#(I37, I38, I39, I40, I41, I42) -> f3#(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] 8) f6#(I43, I44, I45, I46, I47, I48) -> f3#(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] 9) f6#(I49, I50, I51, I52, I53, I54) -> f3#(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] 10) f3#(I55, I56, I57, I58, I59, I60) -> f5#(I55, I56, I57, I58, I59, I60) 11) f4#(I61, I62, I63, I64, I65, I66) -> f1#(I61, I62, I63, I64, I65, I66) We have the following SCCs. { 2, 4, 5, 6, 7, 8, 9, 10 } DP problem for innermost termination. P = f5#(I6, I7, I8, I9, I10, I11) -> f8#(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] f8#(I18, I19, I20, I21, I22, I23) -> f7#(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] f8#(I25, I26, I27, I28, I29, I30) -> f7#(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] f7#(I31, I32, I33, I34, I35, I36) -> f6#(I31, I32, I33, I34, I35, I36) [10 <= I34] f7#(I37, I38, I39, I40, I41, I42) -> f3#(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] f6#(I43, I44, I45, I46, I47, I48) -> f3#(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] f6#(I49, I50, I51, I52, I53, I54) -> f3#(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] f3#(I55, I56, I57, I58, I59, I60) -> f5#(I55, I56, I57, I58, I59, I60) R = f10(x1, x2, x3, x4, x5, x6) -> f9(x1, x2, x3, x4, x5, x6) f9(I0, I1, I2, I3, I4, I5) -> f4(rnd1, 1, 0, rnd4, 1, I5) [rnd4 = 1 /\ rnd1 = 1] f5(I6, I7, I8, I9, I10, I11) -> f8(I6, I7, I8, I9, I10, I11) [1 + I9 <= I6] f5(I12, I13, I14, I15, I16, I17) -> f4(2 + I12, I13, I14, -10 + I15, I16, I17) [I12 <= I15] f8(I18, I19, I20, I21, I22, I23) -> f7(I18, I19, I20, I24, I22, I23) [I24 = I24 /\ 6 <= I21] f8(I25, I26, I27, I28, I29, I30) -> f7(I25, I26, I27, 2 + I28, I29, I30) [I28 <= 5] f7(I31, I32, I33, I34, I35, I36) -> f6(I31, I32, I33, I34, I35, I36) [10 <= I34] f7(I37, I38, I39, I40, I41, I42) -> f3(1 + I37, I38, I39, I40, I41, I42) [1 + I40 <= 10] f6(I43, I44, I45, I46, I47, I48) -> f3(10 + I43, I44, I45, I46, I47, I48) [I46 <= 12] f6(I49, I50, I51, I52, I53, I54) -> f3(1 + I49, I50, I51, I52, I53, I54) [13 <= I52] f3(I55, I56, I57, I58, I59, I60) -> f5(I55, I56, I57, I58, I59, I60) f4(I61, I62, I63, I64, I65, I66) -> f1(I61, I62, I63, I64, I65, I66) f1(I67, I68, I69, I70, I71, I72) -> f3(I67, I68, I69, I70, I71, I72) [1 + I67 <= 30] f1(I73, I74, I75, I76, I77, I78) -> f2(I73, I74, rnd3, I76, I77, 1) [rnd3 = 1 /\ 30 <= I73]