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

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

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

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