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


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YES

DP problem for innermost termination.
P =
  init#(x1, x2) -> f1#(rnd1, rnd2) 
  f2#(I0, I1) -> f2#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
  f1#(I4, I5) -> f2#(I6, I7) [3 <= I6 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I6 - 3 <= I4]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
  f1(I4, I5) -> f2(I6, I7) [3 <= I6 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I6 - 3 <= I4]

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#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
  2) f1#(I4, I5) -> f2#(I6, I7) [3 <= I6 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I6 - 3 <= I4]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  f2#(I0, I1) -> f2#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
  f1(I4, I5) -> f2(I6, I7) [3 <= I6 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I6 - 3 <= I4]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2)] = z1

This gives the following inequalities:
  -1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0  ==>  I0 >! I2

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