/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 =
  sumto#(I0, I1) -> if#(sumto(I0 + 1, I1), I0, I1) [I1 >= I0]
  sumto#(I0, I1) -> sumto#(I0 + 1, I1) [I1 >= I0]
R = 
  if(wrap(z), x, y) -> wrap(x + z) 
  sumto(I0, I1) -> if(sumto(I0 + 1, I1), I0, I1) [I1 >= I0]
  sumto(I2, I3) -> wrap(0) [I2 > I3]

The dependency graph for this problem is:
  0 -> 
  1 -> 0, 1
Where:
  0) sumto#(I0, I1) -> if#(sumto(I0 + 1, I1), I0, I1) [I1 >= I0]
  1) sumto#(I0, I1) -> sumto#(I0 + 1, I1) [I1 >= I0]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  sumto#(I0, I1) -> sumto#(I0 + 1, I1) [I1 >= I0]
R = 
  if(wrap(z), x, y) -> wrap(x + z) 
  sumto(I0, I1) -> if(sumto(I0 + 1, I1), I0, I1) [I1 >= I0]
  sumto(I2, I3) -> wrap(0) [I2 > I3]

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

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

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