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