/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, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
  f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
  f2#(I5, I6) -> f2#(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
  f1#(I7, I8) -> f2#(12, 10) 
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
  f2(I3, I4) -> f2(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
  f2(I5, I6) -> f2(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
  f1(I7, I8) -> f2(12, 10) 

The dependency graph for this problem is:
  0 -> 4
  1 -> 1, 2, 3
  2 -> 1, 2, 3
  3 -> 1, 2
  4 -> 1, 2
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f2#(I0, I1) -> f2#(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
  2) f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
  3) f2#(I5, I6) -> f2#(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
  4) f1#(I7, I8) -> f2#(12, 10) 

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

DP problem for innermost termination.
P =
  f2#(I0, I1) -> f2#(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
  f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
  f2#(I5, I6) -> f2#(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
  f2(I3, I4) -> f2(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
  f2(I5, I6) -> f2(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
  f1(I7, I8) -> f2(12, 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 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1  ==>  I1 >! I1 - 1
  0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1  ==>  I4 (>! \union =) I4
  I5 <= 0 /\ 0 <= I6 - 1  ==>  I6 >! I6 - 1

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
  f2(I3, I4) -> f2(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
  f2(I5, I6) -> f2(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
  f1(I7, I8) -> f2(12, 10) 

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

This gives the following inequalities:
  0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1  ==>  I3 >! I3 - 1

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