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


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  eval#(i, j) -> eval#(i - nat, j + pos) [i - j >= 1 && nat >= 0 && pos > 0]
R = 
  eval(i, j) -> eval(i - nat, j + pos) [i - j >= 1 && nat >= 0 && pos > 0]

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

This gives the following inequalities:
  i - j >= 1 && nat >= 0 && pos > 0  ==>  i - j + -1 * 1 > i - nat - (j + pos) + -1 * 1  with  i - j + -1 * 1 >= 0

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