/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#(x, y, z) -> eval#(x, y + 1, z + 1) [x > y + z]
R = 
  eval(x, y, z) -> eval(x, y + 1, z + 1) [x > y + z]

We use the reverse value criterion with the projection function NU:
  NU[eval#(z1,z2,z3)] = z1 + -1 * (z2 + z3)

This gives the following inequalities:
  x > y + z  ==>  x + -1 * (y + z) > x + -1 * (y + 1 + (z + 1))  with  x + -1 * (y + z) >= 0

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