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

The dependency graph for this problem is:
  0 -> 1
  1 -> 1
Where:
  0) eval#(x, y) -> eval#(y, y) [x > 0 && x > y]
  1) eval#(I0, I1) -> eval#(I0 - 1, I1) [I0 > 0 && I1 >= I0]

We have the following SCCs.
  { 1 }

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

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

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

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