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


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YES

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

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

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

We remove all the strictly oriented dependency pairs.

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

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

This gives the following inequalities:
  I0 > I1 && I0 > I2  ==>  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.