/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, x2) -> f1#(rnd1, rnd2) 
  f3#(I0, I1) -> f2#(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
  f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
  f1#(I5, I6) -> f2#(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
  f2(I3, I4) -> f3(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
  f1(I5, I6) -> f2(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]

The dependency graph for this problem is:
  0 -> 3
  1 -> 2
  2 -> 1
  3 -> 2
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f3#(I0, I1) -> f2#(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
  2) f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
  3) f1#(I5, I6) -> f2#(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f2#(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
  f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
  f2(I3, I4) -> f3(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
  f1(I5, I6) -> f2(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2)] = z2
  NU[f3#(z1,z2)] = z2

This gives the following inequalities:
  0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2  ==>  I1 >! I2
  0 <= I3 - 1 /\ 0 <= I4 - 1  ==>  I4 (>! \union =) I4

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
  f2(I3, I4) -> f3(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
  f1(I5, I6) -> f2(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]

We have the following SCCs.