/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 =
  sumto#(x, y) -> sumto#(x + 1, y) [y >= x]
R = 
  sumto(x, y) -> x + sumto(x + 1, y) [y >= x]
  sumto(I0, I1) -> 0 [I0 > I1]

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

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

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