/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files


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YES

DP problem for innermost termination.
P =
  rand#(I1, y) -> rand#(I1 - 1, id_inc(y)) [I1 > 0]
  rand#(I1, y) -> id_inc#(y) [I1 > 0]
  random#(I4) -> rand#(I4, 0) [I4 >= 0]
R = 
  id_inc(x) -> x + 1 
  id_inc(I0) -> I0 
  rand(I1, y) -> rand(I1 - 1, id_inc(y)) [I1 > 0]
  rand(I2, I3) -> I3 [I2 = 0]
  random(I4) -> rand(I4, 0) [I4 >= 0]

The dependency graph for this problem is:
  0 -> 0, 1
  1 -> 
  2 -> 0, 1
Where:
  0) rand#(I1, y) -> rand#(I1 - 1, id_inc(y)) [I1 > 0]
  1) rand#(I1, y) -> id_inc#(y) [I1 > 0]
  2) random#(I4) -> rand#(I4, 0) [I4 >= 0]

We have the following SCCs.
  { 0 }

DP problem for innermost termination.
P =
  rand#(I1, y) -> rand#(I1 - 1, id_inc(y)) [I1 > 0]
R = 
  id_inc(x) -> x + 1 
  id_inc(I0) -> I0 
  rand(I1, y) -> rand(I1 - 1, id_inc(y)) [I1 > 0]
  rand(I2, I3) -> I3 [I2 = 0]
  random(I4) -> rand(I4, 0) [I4 >= 0]

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

This gives the following inequalities:
  I1 > 0  ==>  I1 > I1 - 1  with  I1 >= 0

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