/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#(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
  f2#(I4, I5) -> f2#(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
  f2#(I8, I9) -> f2#(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
  f1#(I12, I13) -> f2#(I14, I15) [4 <= I14 - 1 /\ 0 <= I15 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
  f2(I4, I5) -> f2(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
  f2(I8, I9) -> f2(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
  f1(I12, I13) -> f2(I14, I15) [4 <= I14 - 1 /\ 0 <= I15 - 1]

The dependency graph for this problem is:
  0 -> 4
  1 -> 1, 2, 3
  2 -> 1, 2, 3
  3 -> 1, 2, 3
  4 -> 1, 2, 3
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f2#(I0, I1) -> f2#(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
  2) f2#(I4, I5) -> f2#(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
  3) f2#(I8, I9) -> f2#(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
  4) f1#(I12, I13) -> f2#(I14, I15) [4 <= I14 - 1 /\ 0 <= I15 - 1]

We have the following SCCs.
  { 1, 2, 3 }

DP problem for innermost termination.
P =
  f2#(I0, I1) -> f2#(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
  f2#(I4, I5) -> f2#(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
  f2#(I8, I9) -> f2#(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
  f2(I4, I5) -> f2(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
  f2(I8, I9) -> f2(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
  f1(I12, I13) -> f2(I14, I15) [4 <= I14 - 1 /\ 0 <= I15 - 1]

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

This gives the following inequalities:
  -1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0  ==>  I0 >! I2
  -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4  ==>  I4 >! I6
  -1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8  ==>  I8 >! I10

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