/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_2#(x, y) -> eval_1#(x, y) [0 >= y]
  eval_2#(I0, I1) -> eval_2#(I0 - 1, I1 - 1) [I1 > 0]
  eval_1#(I2, I3) -> eval_2#(I2, I3) [I2 = I3 && I2 > 0]
R = 
  eval_2(x, y) -> eval_1(x, y) [0 >= y]
  eval_2(I0, I1) -> eval_2(I0 - 1, I1 - 1) [I1 > 0]
  eval_1(I2, I3) -> eval_2(I2, I3) [I2 = I3 && I2 > 0]

The dependency graph for this problem is:
  0 -> 
  1 -> 0, 1
  2 -> 1
Where:
  0) eval_2#(x, y) -> eval_1#(x, y) [0 >= y]
  1) eval_2#(I0, I1) -> eval_2#(I0 - 1, I1 - 1) [I1 > 0]
  2) eval_1#(I2, I3) -> eval_2#(I2, I3) [I2 = I3 && I2 > 0]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  eval_2#(I0, I1) -> eval_2#(I0 - 1, I1 - 1) [I1 > 0]
R = 
  eval_2(x, y) -> eval_1(x, y) [0 >= y]
  eval_2(I0, I1) -> eval_2(I0 - 1, I1 - 1) [I1 > 0]
  eval_1(I2, I3) -> eval_2(I2, I3) [I2 = I3 && I2 > 0]

We use the reverse value criterion with the projection function NU:
  NU[eval_2#(z1,z2)] = z2 + -1 * 0

This gives the following inequalities:
  I1 > 0  ==>  I1 + -1 * 0 > I1 - 1 + -1 * 0  with  I1 + -1 * 0 >= 0

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