/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  init#(x1) -> f1#(rnd1) 
  f2#(I0) -> f2#(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1 /\ I1 + 1 <= I0]
  f1#(I2) -> f2#(I3) [3 <= I3 - 1]
R = 
  init(x1) -> f1(rnd1) 
  f2(I0) -> f2(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1 /\ I1 + 1 <= I0]
  f1(I2) -> f2(I3) [3 <= I3 - 1]

The dependency graph for this problem is:
  0 -> 2
  1 -> 1
  2 -> 1
Where:
  0) init#(x1) -> f1#(rnd1) 
  1) f2#(I0) -> f2#(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1 /\ I1 + 1 <= I0]
  2) f1#(I2) -> f2#(I3) [3 <= I3 - 1]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  f2#(I0) -> f2#(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1 /\ I1 + 1 <= I0]
R = 
  init(x1) -> f1(rnd1) 
  f2(I0) -> f2(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1 /\ I1 + 1 <= I0]
  f1(I2) -> f2(I3) [3 <= I3 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1)] = z1

This gives the following inequalities:
  -1 <= I1 - 1 /\ 0 <= I0 - 1 /\ I1 + 1 <= I0  ==>  I0 >! I1

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.