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


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


YES

DP problem for innermost termination.
P =
  init#(x1, x2) -> f1#(rnd1, rnd2) 
  f2#(I0, I1) -> f2#(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
  f2#(I2, I3) -> f2#(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
  f1#(I4, I5) -> f2#(0, 10) 
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
  f2(I2, I3) -> f2(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
  f1(I4, I5) -> f2(0, 10) 

The dependency graph for this problem is:
  0 -> 3
  1 -> 1, 2
  2 -> 1, 2
  3 -> 1, 2
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f2#(I0, I1) -> f2#(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
  2) f2#(I2, I3) -> f2#(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
  3) f1#(I4, I5) -> f2#(0, 10) 

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f2#(I0, I1) -> f2#(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
  f2#(I2, I3) -> f2#(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
  f2(I2, I3) -> f2(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
  f1(I4, I5) -> f2(0, 10) 

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

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

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I0, I1) -> f2#(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
  f2(I2, I3) -> f2(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
  f1(I4, I5) -> f2(0, 10) 

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

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

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