/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#(x, y, z) -> eval#(x, y, z + 1) [y > x && x >= z]
  eval#(I0, I1, I2) -> eval#(I0 + 1, I1, I2) [I1 > I0 && I2 > I0]
R = 
  eval(x, y, z) -> eval(x, y, z + 1) [y > x && x >= z]
  eval(I0, I1, I2) -> eval(I0 + 1, I1, I2) [I1 > I0 && I2 > I0]

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

This gives the following inequalities:
  y > x && x >= z  ==>  y + -1 * x >= y + -1 * x
  I1 > I0 && I2 > I0  ==>  I1 + -1 * I0 > I1 + -1 * (I0 + 1)  with  I1 + -1 * I0 >= 0

We remove all the strictly oriented dependency pairs.

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

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

This gives the following inequalities:
  y > x && x >= z  ==>  x + -1 * z > x + -1 * (z + 1)  with  x + -1 * z >= 0

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