/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 =
  eval#(l, u) -> eval#(l, (l + u) / 2) [l >= 0 && u >= 0 && l < u]
  eval#(I0, I1) -> eval#((I0 + I1) / 2 + 1, I1) [I0 >= 0 && I1 >= 0 && I0 < I1]
R = 
  eval(l, u) -> eval(l, (l + u) / 2) [l >= 0 && u >= 0 && l < u]
  eval(I0, I1) -> eval((I0 + I1) / 2 + 1, I1) [I0 >= 0 && I1 >= 0 && I0 < I1]

We use the reverse value criterion with the projection function NU:
  NU[eval#(z1,z2)] = z2 + -1 * z1

This gives the following inequalities:
  l >= 0 && u >= 0 && l < u  ==>  u + -1 * l > (l + u) / 2 + -1 * l  with  u + -1 * l >= 0
  I0 >= 0 && I1 >= 0 && I0 < I1  ==>  I1 + -1 * I0 > I1 + -1 * ((I0 + I1) / 2 + 1)  with  I1 + -1 * I0 >= 0

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