/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 =
  gcd#(x, y) -> gcd#(y - x, x) [y > x && x > 0]
  gcd#(I0, I1) -> gcd#(I0 - I1, I1) [I0 >= I1 && I1 > 0]
R = 
  gcd(x, y) -> gcd(y - x, x) [y > x && x > 0]
  gcd(I0, I1) -> gcd(I0 - I1, I1) [I0 >= I1 && I1 > 0]
  gcd(0, I2) -> I2 
  gcd(I3, 0) -> I3 

We use the reverse value criterion with the projection function NU:
  NU[gcd#(z1,z2)] = z2

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

We remove all the strictly oriented dependency pairs.

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

We use the reverse value criterion with the projection function NU:
  NU[gcd#(z1,z2)] = z1

This gives the following inequalities:
  I0 >= I1 && I1 > 0  ==>  I0 > I0 - I1  with  I0 >= 0

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