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


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YES

DP problem for innermost termination.
P =
  eval#(x, y) -> eval#(x, y - 1) [x > 0 && y > 0]
  eval#(I0, I1) -> eval#(I0 - 1, z) [I0 > 0 && I1 > 0]
R = 
  eval(x, y) -> eval(x, y - 1) [x > 0 && y > 0]
  eval(I0, I1) -> eval(I0 - 1, z) [I0 > 0 && I1 > 0]

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

This gives the following inequalities:
  x > 0 && y > 0  ==>  x >= x
  I0 > 0 && I1 > 0  ==>  I0 > I0 - 1  with  I0 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  eval#(x, y) -> eval#(x, y - 1) [x > 0 && y > 0]
R = 
  eval(x, y) -> eval(x, y - 1) [x > 0 && y > 0]
  eval(I0, I1) -> eval(I0 - 1, z) [I0 > 0 && I1 > 0]

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

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

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