/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 =
  div#(x, y) -> div#(x - y, y) [x > y && y > 0]
R = 
  div(x, y) -> div(x - y, y) + 1 [x > y && y > 0]
  div(I0, I1) -> 0 [I1 >= I0]
  div(I2, I3) -> 0 [0 >= I3]

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

This gives the following inequalities:
  x > y && y > 0  ==>  x > x - y  with  x >= 0

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