/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  eval_2#(x, y) -> eval_1#(x - 1, y) [x > 0 && y >= 0 && y >= x]
  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
  eval_1#(I2, I3) -> eval_2#(I2, 0) [I2 > 0]
R = 
  eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && y >= 0 && y >= x]
  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
  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
  NU[eval_2#(z1,z2)] = z1

This gives the following inequalities:
  x > 0 && y >= 0 && y >= x  ==>  x > x - 1  with  x >= 0
  I0 > 0 && I1 >= 0 && I0 > I1  ==>  I0 >= I0
  I2 > 0  ==>  I2 >= I2

We remove all the strictly oriented dependency pairs.

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

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

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
R = 
  eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && y >= 0 && y >= x]
  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
  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 * z2

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

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