/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) f2#(I0, I1) -> f2#(I0 - 1, I2) [I0 <= 0 /\ I0 <= 1] f2#(I3, I4) -> f2#(I3 + 1, I5) [0 <= I3 - 1] f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 - 1, I2) [I0 <= 0 /\ I0 <= 1] f2(I3, I4) -> f2(I3 + 1, I5) [0 <= I3 - 1] f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1] The dependency graph for this problem is: 0 -> 3 1 -> 1 2 -> 2 3 -> 1, 2 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f2#(I0, I1) -> f2#(I0 - 1, I2) [I0 <= 0 /\ I0 <= 1] 2) f2#(I3, I4) -> f2#(I3 + 1, I5) [0 <= I3 - 1] 3) f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1] We have the following SCCs. { 2 } { 1 } DP problem for innermost termination. P = f2#(I0, I1) -> f2#(I0 - 1, I2) [I0 <= 0 /\ I0 <= 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 - 1, I2) [I0 <= 0 /\ I0 <= 1] f2(I3, I4) -> f2(I3 + 1, I5) [0 <= I3 - 1] f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]