/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 = f6#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) f5#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) f2#(I5, I6, I7, I8, I9) -> f5#(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] f3#(I15, I16, I17, I18, I19) -> f2#(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1#(I20, I21, I22, I23, I24) -> f2#(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0] R = f6(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] f2(I10, I11, I12, I13, I14) -> f4(I10, I11, I12, I14, I14) [10 <= I11] f3(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1(I20, I21, I22, I23, I24) -> f2(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0] The dependency graph for this problem is: 0 -> 4 1 -> 2 2 -> 1 3 -> 2 4 -> 2 Where: 0) f6#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 1) f5#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 2) f2#(I5, I6, I7, I8, I9) -> f5#(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] 3) f3#(I15, I16, I17, I18, I19) -> f2#(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] 4) f1#(I20, I21, I22, I23, I24) -> f2#(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f5#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) f2#(I5, I6, I7, I8, I9) -> f5#(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] R = f6(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] f2(I10, I11, I12, I13, I14) -> f4(I10, I11, I12, I14, I14) [10 <= I11] f3(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1(I20, I21, I22, I23, I24) -> f2(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0]