/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) -> f1#(rnd1) 
  f2#(I0) -> f2#(I0 + 1) [I0 <= 10]
  f1#(I1) -> f2#(0) 
R = 
  init(x1) -> f1(rnd1) 
  f2(I0) -> f2(I0 + 1) [I0 <= 10]
  f1(I1) -> f2(0) 

The dependency graph for this problem is:
  0 -> 2
  1 -> 1
  2 -> 1
Where:
  0) init#(x1) -> f1#(rnd1) 
  1) f2#(I0) -> f2#(I0 + 1) [I0 <= 10]
  2) f1#(I1) -> f2#(0) 

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  f2#(I0) -> f2#(I0 + 1) [I0 <= 10]
R = 
  init(x1) -> f1(rnd1) 
  f2(I0) -> f2(I0 + 1) [I0 <= 10]
  f1(I1) -> f2(0) 

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1)] = 10 + -1 * z1

This gives the following inequalities:
  I0 <= 10  ==>  10 + -1 * I0 > 10 + -1 * (I0 + 1)  with  10 + -1 * I0 >= 0

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