/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 =
  f#(I0, I1, I2, I3) -> f#(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
  sqrt#(I4) -> f#(I4, 0, 1, 1) 
R = 
  f(x, z, u, w) -> z [w > x]
  f(I0, I1, I2, I3) -> f(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
  sqrt(I4) -> f(I4, 0, 1, 1) 

The dependency graph for this problem is:
  0 -> 0
  1 -> 0
Where:
  0) f#(I0, I1, I2, I3) -> f#(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
  1) sqrt#(I4) -> f#(I4, 0, 1, 1) 

We have the following SCCs.
  { 0 }

DP problem for innermost termination.
P =
  f#(I0, I1, I2, I3) -> f#(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
R = 
  f(x, z, u, w) -> z [w > x]
  f(I0, I1, I2, I3) -> f(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
  sqrt(I4) -> f(I4, 0, 1, 1) 

We use the reverse value criterion with the projection function NU:
  NU[f#(z1,z2,z3,z4)] = z1 + -1 * z4

This gives the following inequalities:
  I0 >= I3 && I2 >= 0  ==>  I0 + -1 * I3 > I0 + -1 * (I3 + I2 + 2)  with  I0 + -1 * I3 >= 0

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