/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f3#(I0 - 1, I2) [0 <= I0 - 1] f3#(I3, I4) -> f3#(I3 - 1, I5) [I3 <= -1] f2#(I6, I7) -> f3#(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8] f1#(I10, I11) -> f2#(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ I13 - 2 * I14 = 0 /\ -1 <= I13 - 1 /\ 0 <= I10 - 1] f2#(I15, I16) -> f3#(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ I19 - 2 * I20 = 1 /\ -1 <= I19 - 1 /\ 0 <= I15 - 1 /\ I19 - 2 * I20 <= 1 /\ 0 <= I19 - 2 * I20] f1#(I21, I22) -> f2#(I21, I22) [-1 <= I23 - 1 /\ 1 <= I22 - 1 /\ I24 - 2 * I25 = 1 /\ -1 <= I24 - 1 /\ 0 <= I21 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f3(I0 - 1, I2) [0 <= I0 - 1] f3(I3, I4) -> f3(I3 - 1, I5) [I3 <= -1] f2(I6, I7) -> f3(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8] f1(I10, I11) -> f2(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ I13 - 2 * I14 = 0 /\ -1 <= I13 - 1 /\ 0 <= I10 - 1] f2(I15, I16) -> f3(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ I19 - 2 * I20 = 1 /\ -1 <= I19 - 1 /\ 0 <= I15 - 1 /\ I19 - 2 * I20 <= 1 /\ 0 <= I19 - 2 * I20] f1(I21, I22) -> f2(I21, I22) [-1 <= I23 - 1 /\ 1 <= I22 - 1 /\ I24 - 2 * I25 = 1 /\ -1 <= I24 - 1 /\ 0 <= I21 - 1] The dependency graph for this problem is: 0 -> 4, 6 1 -> 1 2 -> 2 3 -> 2 4 -> 3, 5 5 -> 1 6 -> 3, 5 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f3#(I0 - 1, I2) [0 <= I0 - 1] 2) f3#(I3, I4) -> f3#(I3 - 1, I5) [I3 <= -1] 3) f2#(I6, I7) -> f3#(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8] 4) f1#(I10, I11) -> f2#(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ I13 - 2 * I14 = 0 /\ -1 <= I13 - 1 /\ 0 <= I10 - 1] 5) f2#(I15, I16) -> f3#(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ I19 - 2 * I20 = 1 /\ -1 <= I19 - 1 /\ 0 <= I15 - 1 /\ I19 - 2 * I20 <= 1 /\ 0 <= I19 - 2 * I20] 6) f1#(I21, I22) -> f2#(I21, I22) [-1 <= I23 - 1 /\ 1 <= I22 - 1 /\ I24 - 2 * I25 = 1 /\ -1 <= I24 - 1 /\ 0 <= I21 - 1] We have the following SCCs. { 1 } { 2 } DP problem for innermost termination. P = f3#(I3, I4) -> f3#(I3 - 1, I5) [I3 <= -1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f3(I0 - 1, I2) [0 <= I0 - 1] f3(I3, I4) -> f3(I3 - 1, I5) [I3 <= -1] f2(I6, I7) -> f3(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8] f1(I10, I11) -> f2(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ I13 - 2 * I14 = 0 /\ -1 <= I13 - 1 /\ 0 <= I10 - 1] f2(I15, I16) -> f3(I17, I18) [-1 <= I17 - 1 /\ 1 <= I16 - 1 /\ I19 - 2 * I20 = 1 /\ -1 <= I19 - 1 /\ 0 <= I15 - 1 /\ I19 - 2 * I20 <= 1 /\ 0 <= I19 - 2 * I20] f1(I21, I22) -> f2(I21, I22) [-1 <= I23 - 1 /\ 1 <= I22 - 1 /\ I24 - 2 * I25 = 1 /\ -1 <= I24 - 1 /\ 0 <= I21 - 1]