/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 =
  log#(x, y) -> log#((x - y) / y, y) [x >= 2 && y >= 2]
R = 
  log(x, y) -> 1 + log((x - y) / y, y) [x >= 2 && y >= 2]
  log(1, I0) -> 0 [I0 >= 2]

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

This gives the following inequalities:
  x >= 2 && y >= 2  ==>  x > (x - y) / y  with  x >= 0

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