/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 =
  eval#(x) -> eval#(x / 2) [x > 0 && not(x = 0) && x % 2 = 0]
  eval#(I0) -> eval#(I0 - 1) [I0 > 0 && not(I0 = 0) && I0 % 2 > 0]
R = 
  eval(x) -> eval(x / 2) [x > 0 && not(x = 0) && x % 2 = 0]
  eval(I0) -> eval(I0 - 1) [I0 > 0 && not(I0 = 0) && I0 % 2 > 0]

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

This gives the following inequalities:
  x > 0 && not(x = 0) && x % 2 = 0  ==>  x > x / 2  with  x >= 0
  I0 > 0 && not(I0 = 0) && I0 % 2 > 0  ==>  I0 > I0 - 1  with  I0 >= 0

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