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


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YES

DP problem for innermost termination.
P =
  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  f2#(I0, I1, I2) -> f2#(I3, I1 + 1, I2) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 <= I0 /\ -1 <= I2 - 1 /\ I1 <= I2 - 1]
  f1#(I4, I5, I6) -> f2#(I7, 0, I5) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I7 <= I4]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(I3, I1 + 1, I2) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 <= I0 /\ -1 <= I2 - 1 /\ I1 <= I2 - 1]
  f1(I4, I5, I6) -> f2(I7, 0, I5) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I7 <= I4]

The dependency graph for this problem is:
  0 -> 2
  1 -> 1
  2 -> 1
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f2#(I0, I1, I2) -> f2#(I3, I1 + 1, I2) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 <= I0 /\ -1 <= I2 - 1 /\ I1 <= I2 - 1]
  2) f1#(I4, I5, I6) -> f2#(I7, 0, I5) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I7 <= I4]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  f2#(I0, I1, I2) -> f2#(I3, I1 + 1, I2) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 <= I0 /\ -1 <= I2 - 1 /\ I1 <= I2 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(I3, I1 + 1, I2) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 <= I0 /\ -1 <= I2 - 1 /\ I1 <= I2 - 1]
  f1(I4, I5, I6) -> f2(I7, 0, I5) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I7 <= I4]

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z3 - 1 + -1 * z2

This gives the following inequalities:
  0 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 <= I0 /\ -1 <= I2 - 1 /\ I1 <= I2 - 1  ==>  I2 - 1 + -1 * I1 > I2 - 1 + -1 * (I1 + 1)  with  I2 - 1 + -1 * I1 >= 0

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