/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 = init#(x1, x2) -> f1#(rnd1, rnd2) f5#(I0, I1) -> f3#(I0 + 1, I2) [I0 <= -1] f5#(I3, I4) -> f4#(I3, I5) [-1 <= I3 - 1] f3#(I6, I7) -> f5#(I6, I8) [-6 <= I6 - 1 /\ I6 <= 5 /\ I6 <= 0] f4#(I9, I10) -> f5#(I9 - 1, I11) [I9 <= 5 /\ 0 <= I9 - 1] f4#(I12, I13) -> f5#(0, I14) [0 = I12] f2#(I15, I16) -> f4#(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ y1 - 2 * y2 = 1 /\ -1 <= y1 - 1 /\ 0 <= I15 - 1 /\ y1 - 2 * y2 <= 1 /\ 0 <= y1 - 2 * y2] f1#(I19, I20) -> f2#(I19, I20) [-1 <= I21 - 1 /\ 1 <= I20 - 1 /\ I22 - 2 * y3 = 1 /\ -1 <= I22 - 1 /\ 0 <= I19 - 1] f2#(I23, I24) -> f3#(I25, I26) [-1 <= I27 - 1 /\ 1 <= I24 - 1 /\ I28 - 2 * I29 = 0 /\ -1 <= I28 - 1 /\ 0 <= I23 - 1 /\ I28 - 2 * I29 <= 1 /\ 0 <= I28 - 2 * I29 /\ 0 - I27 = I25] f1#(I30, I31) -> f2#(I30, I31) [-1 <= I32 - 1 /\ 1 <= I31 - 1 /\ I33 - 2 * I34 = 0 /\ -1 <= I33 - 1 /\ 0 <= I30 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f5(I0, I1) -> f3(I0 + 1, I2) [I0 <= -1] f5(I3, I4) -> f4(I3, I5) [-1 <= I3 - 1] f3(I6, I7) -> f5(I6, I8) [-6 <= I6 - 1 /\ I6 <= 5 /\ I6 <= 0] f4(I9, I10) -> f5(I9 - 1, I11) [I9 <= 5 /\ 0 <= I9 - 1] f4(I12, I13) -> f5(0, I14) [0 = I12] f2(I15, I16) -> f4(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ y1 - 2 * y2 = 1 /\ -1 <= y1 - 1 /\ 0 <= I15 - 1 /\ y1 - 2 * y2 <= 1 /\ 0 <= y1 - 2 * y2] f1(I19, I20) -> f2(I19, I20) [-1 <= I21 - 1 /\ 1 <= I20 - 1 /\ I22 - 2 * y3 = 1 /\ -1 <= I22 - 1 /\ 0 <= I19 - 1] f2(I23, I24) -> f3(I25, I26) [-1 <= I27 - 1 /\ 1 <= I24 - 1 /\ I28 - 2 * I29 = 0 /\ -1 <= I28 - 1 /\ 0 <= I23 - 1 /\ I28 - 2 * I29 <= 1 /\ 0 <= I28 - 2 * I29 /\ 0 - I27 = I25] f1(I30, I31) -> f2(I30, I31) [-1 <= I32 - 1 /\ 1 <= I31 - 1 /\ I33 - 2 * I34 = 0 /\ -1 <= I33 - 1 /\ 0 <= I30 - 1] The dependency graph for this problem is: 0 -> 7, 9 1 -> 3 2 -> 4, 5 3 -> 1, 2 4 -> 2 5 -> 2 6 -> 4, 5 7 -> 6, 8 8 -> 3 9 -> 6, 8 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f5#(I0, I1) -> f3#(I0 + 1, I2) [I0 <= -1] 2) f5#(I3, I4) -> f4#(I3, I5) [-1 <= I3 - 1] 3) f3#(I6, I7) -> f5#(I6, I8) [-6 <= I6 - 1 /\ I6 <= 5 /\ I6 <= 0] 4) f4#(I9, I10) -> f5#(I9 - 1, I11) [I9 <= 5 /\ 0 <= I9 - 1] 5) f4#(I12, I13) -> f5#(0, I14) [0 = I12] 6) f2#(I15, I16) -> f4#(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ y1 - 2 * y2 = 1 /\ -1 <= y1 - 1 /\ 0 <= I15 - 1 /\ y1 - 2 * y2 <= 1 /\ 0 <= y1 - 2 * y2] 7) f1#(I19, I20) -> f2#(I19, I20) [-1 <= I21 - 1 /\ 1 <= I20 - 1 /\ I22 - 2 * y3 = 1 /\ -1 <= I22 - 1 /\ 0 <= I19 - 1] 8) f2#(I23, I24) -> f3#(I25, I26) [-1 <= I27 - 1 /\ 1 <= I24 - 1 /\ I28 - 2 * I29 = 0 /\ -1 <= I28 - 1 /\ 0 <= I23 - 1 /\ I28 - 2 * I29 <= 1 /\ 0 <= I28 - 2 * I29 /\ 0 - I27 = I25] 9) f1#(I30, I31) -> f2#(I30, I31) [-1 <= I32 - 1 /\ 1 <= I31 - 1 /\ I33 - 2 * I34 = 0 /\ -1 <= I33 - 1 /\ 0 <= I30 - 1] We have the following SCCs. { 1, 3 } { 2, 4, 5 } DP problem for innermost termination. P = f5#(I3, I4) -> f4#(I3, I5) [-1 <= I3 - 1] f4#(I9, I10) -> f5#(I9 - 1, I11) [I9 <= 5 /\ 0 <= I9 - 1] f4#(I12, I13) -> f5#(0, I14) [0 = I12] R = init(x1, x2) -> f1(rnd1, rnd2) f5(I0, I1) -> f3(I0 + 1, I2) [I0 <= -1] f5(I3, I4) -> f4(I3, I5) [-1 <= I3 - 1] f3(I6, I7) -> f5(I6, I8) [-6 <= I6 - 1 /\ I6 <= 5 /\ I6 <= 0] f4(I9, I10) -> f5(I9 - 1, I11) [I9 <= 5 /\ 0 <= I9 - 1] f4(I12, I13) -> f5(0, I14) [0 = I12] f2(I15, I16) -> f4(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ y1 - 2 * y2 = 1 /\ -1 <= y1 - 1 /\ 0 <= I15 - 1 /\ y1 - 2 * y2 <= 1 /\ 0 <= y1 - 2 * y2] f1(I19, I20) -> f2(I19, I20) [-1 <= I21 - 1 /\ 1 <= I20 - 1 /\ I22 - 2 * y3 = 1 /\ -1 <= I22 - 1 /\ 0 <= I19 - 1] f2(I23, I24) -> f3(I25, I26) [-1 <= I27 - 1 /\ 1 <= I24 - 1 /\ I28 - 2 * I29 = 0 /\ -1 <= I28 - 1 /\ 0 <= I23 - 1 /\ I28 - 2 * I29 <= 1 /\ 0 <= I28 - 2 * I29 /\ 0 - I27 = I25] f1(I30, I31) -> f2(I30, I31) [-1 <= I32 - 1 /\ 1 <= I31 - 1 /\ I33 - 2 * I34 = 0 /\ -1 <= I33 - 1 /\ 0 <= I30 - 1] We use the basic value criterion with the projection function NU: NU[f4#(z1,z2)] = z1 NU[f5#(z1,z2)] = z1 This gives the following inequalities: -1 <= I3 - 1 ==> I3 (>! \union =) I3 I9 <= 5 /\ 0 <= I9 - 1 ==> I9 >! I9 - 1 0 = I12 ==> I12 (>! \union =) 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I3, I4) -> f4#(I3, I5) [-1 <= I3 - 1] f4#(I12, I13) -> f5#(0, I14) [0 = I12] R = init(x1, x2) -> f1(rnd1, rnd2) f5(I0, I1) -> f3(I0 + 1, I2) [I0 <= -1] f5(I3, I4) -> f4(I3, I5) [-1 <= I3 - 1] f3(I6, I7) -> f5(I6, I8) [-6 <= I6 - 1 /\ I6 <= 5 /\ I6 <= 0] f4(I9, I10) -> f5(I9 - 1, I11) [I9 <= 5 /\ 0 <= I9 - 1] f4(I12, I13) -> f5(0, I14) [0 = I12] f2(I15, I16) -> f4(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ y1 - 2 * y2 = 1 /\ -1 <= y1 - 1 /\ 0 <= I15 - 1 /\ y1 - 2 * y2 <= 1 /\ 0 <= y1 - 2 * y2] f1(I19, I20) -> f2(I19, I20) [-1 <= I21 - 1 /\ 1 <= I20 - 1 /\ I22 - 2 * y3 = 1 /\ -1 <= I22 - 1 /\ 0 <= I19 - 1] f2(I23, I24) -> f3(I25, I26) [-1 <= I27 - 1 /\ 1 <= I24 - 1 /\ I28 - 2 * I29 = 0 /\ -1 <= I28 - 1 /\ 0 <= I23 - 1 /\ I28 - 2 * I29 <= 1 /\ 0 <= I28 - 2 * I29 /\ 0 - I27 = I25] f1(I30, I31) -> f2(I30, I31) [-1 <= I32 - 1 /\ 1 <= I31 - 1 /\ I33 - 2 * I34 = 0 /\ -1 <= I33 - 1 /\ 0 <= I30 - 1]