/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#(i, j) -> eval_1#(i - 1, j) [j > i - 1]
  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1]
  eval_1#(I2, I3) -> eval_2#(I2, 0) [I2 >= 0]
R = 
  eval_2(i, j) -> eval_1(i - 1, j) [j > i - 1]
  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I1 <= I0 - 1]
  eval_1(I2, I3) -> eval_2(I2, 0) [I2 >= 0]

We use the reverse value criterion with the projection function NU:
  NU[eval_1#(z1,z2)] = z1 + -1 * 0
  NU[eval_2#(z1,z2)] = z1 - 1 + -1 * 0

This gives the following inequalities:
  j > i - 1  ==>  i - 1 + -1 * 0 >= i - 1 + -1 * 0
  I1 <= I0 - 1  ==>  I0 - 1 + -1 * 0 >= I0 - 1 + -1 * 0
  I2 >= 0  ==>  I2 + -1 * 0 > I2 - 1 + -1 * 0  with  I2 + -1 * 0 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  eval_2#(i, j) -> eval_1#(i - 1, j) [j > i - 1]
  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1]
R = 
  eval_2(i, j) -> eval_1(i - 1, j) [j > i - 1]
  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I1 <= I0 - 1]
  eval_1(I2, I3) -> eval_2(I2, 0) [I2 >= 0]

The dependency graph for this problem is:
  0 -> 
  1 -> 0, 1
Where:
  0) eval_2#(i, j) -> eval_1#(i - 1, j) [j > i - 1]
  1) eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1]
R = 
  eval_2(i, j) -> eval_1(i - 1, j) [j > i - 1]
  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I1 <= I0 - 1]
  eval_1(I2, I3) -> eval_2(I2, 0) [I2 >= 0]

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

This gives the following inequalities:
  I1 <= I0 - 1  ==>  I0 - 1 + -1 * I1 > I0 - 1 + -1 * (I1 + 1)  with  I0 - 1 + -1 * I1 >= 0

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