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

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

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

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