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