/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files


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MAYBE

DP problem for innermost termination.
P =
  init#(x1, x2) -> f1#(rnd1, rnd2) 
  f2#(I0, I1) -> f2#(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
  f1#(I2, I3) -> f2#(I3 + 5, I3) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I3 + 5 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
  f1(I2, I3) -> f2(I3 + 5, I3) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I3 + 5 - 1]

The dependency graph for this problem is:
  0 -> 2
  1 -> 1
  2 -> 1
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f2#(I0, I1) -> f2#(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
  2) f1#(I2, I3) -> f2#(I3 + 5, I3) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I3 + 5 - 1]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  f2#(I0, I1) -> f2#(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
  f1(I2, I3) -> f2(I3 + 5, I3) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I3 + 5 - 1]