/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 = init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) f4#(I0, I1, I2) -> f4#(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0] f4#(I5, I6, I7) -> f4#(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5] f4#(I11, I12, I13) -> f4#(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11] f3#(I17, I18, I19) -> f4#(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19] f2#(I22, I23, I24) -> f2#(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1] f1#(I27, I28, I29) -> f3#(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1] f1#(I33, I34, I35) -> f3#(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1] f1#(I39, I40, I41) -> f2#(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0] f4(I5, I6, I7) -> f4(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5] f4(I11, I12, I13) -> f4(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11] f3(I17, I18, I19) -> f4(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19] f2(I22, I23, I24) -> f2(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1] f1(I27, I28, I29) -> f3(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1] f1(I33, I34, I35) -> f3(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1] f1(I39, I40, I41) -> f2(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1] The dependency graph for this problem is: 0 -> 6, 7, 8 1 -> 2 2 -> 1, 2, 3 3 -> 1, 2, 3 4 -> 1 5 -> 5 6 -> 4 7 -> 4 8 -> 5 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f4#(I0, I1, I2) -> f4#(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0] 2) f4#(I5, I6, I7) -> f4#(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5] 3) f4#(I11, I12, I13) -> f4#(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11] 4) f3#(I17, I18, I19) -> f4#(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19] 5) f2#(I22, I23, I24) -> f2#(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1] 6) f1#(I27, I28, I29) -> f3#(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1] 7) f1#(I33, I34, I35) -> f3#(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1] 8) f1#(I39, I40, I41) -> f2#(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1] We have the following SCCs. { 1, 2, 3 } { 5 } DP problem for innermost termination. P = f2#(I22, I23, I24) -> f2#(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0] f4(I5, I6, I7) -> f4(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5] f4(I11, I12, I13) -> f4(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11] f3(I17, I18, I19) -> f4(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19] f2(I22, I23, I24) -> f2(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1] f1(I27, I28, I29) -> f3(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1] f1(I33, I34, I35) -> f3(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1] f1(I39, I40, I41) -> f2(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = z1 This gives the following inequalities: I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1 ==> I22 >! I22 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f4#(I0, I1, I2) -> f4#(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0] f4#(I5, I6, I7) -> f4#(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5] f4#(I11, I12, I13) -> f4#(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0] f4(I5, I6, I7) -> f4(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5] f4(I11, I12, I13) -> f4(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11] f3(I17, I18, I19) -> f4(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19] f2(I22, I23, I24) -> f2(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1] f1(I27, I28, I29) -> f3(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1] f1(I33, I34, I35) -> f3(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1] f1(I39, I40, I41) -> f2(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1] We use the basic value criterion with the projection function NU: NU[f4#(z1,z2,z3)] = z2 This gives the following inequalities: 0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0 ==> I1 (>! \union =) I4 I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5 ==> I6 >! I9 I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11 ==> I12 >! I15 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I0, I1, I2) -> f4#(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0] f4(I5, I6, I7) -> f4(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5] f4(I11, I12, I13) -> f4(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11] f3(I17, I18, I19) -> f4(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19] f2(I22, I23, I24) -> f2(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1] f1(I27, I28, I29) -> f3(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1] f1(I33, I34, I35) -> f3(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1] f1(I39, I40, I41) -> f2(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1] The dependency graph for this problem is: 1 -> Where: 1) f4#(I0, I1, I2) -> f4#(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0] We have the following SCCs.