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