/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, z) -> eval_1#(x, y, z) [z >= y]
  eval_2#(I0, I1, I2) -> eval_2#(I0 - 1, I1 - 1, I2) [I1 > I2]
  eval_1#(I3, I4, I5) -> eval_2#(I3, I4, I5) [I3 = I4 && I3 > I5]
R = 
  eval_2(x, y, z) -> eval_1(x, y, z) [z >= y]
  eval_2(I0, I1, I2) -> eval_2(I0 - 1, I1 - 1, I2) [I1 > I2]
  eval_1(I3, I4, I5) -> eval_2(I3, I4, I5) [I3 = I4 && I3 > I5]

The dependency graph for this problem is:
  0 -> 
  1 -> 0, 1
  2 -> 1
Where:
  0) eval_2#(x, y, z) -> eval_1#(x, y, z) [z >= y]
  1) eval_2#(I0, I1, I2) -> eval_2#(I0 - 1, I1 - 1, I2) [I1 > I2]
  2) eval_1#(I3, I4, I5) -> eval_2#(I3, I4, I5) [I3 = I4 && I3 > I5]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  eval_2#(I0, I1, I2) -> eval_2#(I0 - 1, I1 - 1, I2) [I1 > I2]
R = 
  eval_2(x, y, z) -> eval_1(x, y, z) [z >= y]
  eval_2(I0, I1, I2) -> eval_2(I0 - 1, I1 - 1, I2) [I1 > I2]
  eval_1(I3, I4, I5) -> eval_2(I3, I4, I5) [I3 = I4 && I3 > I5]

We use the reverse value criterion with the projection function NU:
  NU[eval_2#(z1,z2,z3)] = z2 + -1 * z3

This gives the following inequalities:
  I1 > I2  ==>  I1 + -1 * I2 > I1 - 1 + -1 * I2  with  I1 + -1 * I2 >= 0

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