/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 =
  sif#(true, I0, I1) -> sum#(I0, I1 + 1) 
  sum#(I2, I3) -> sif#(I2 >= I3, I2, I3) 
R = 
  sif(false, x, y) -> 0 
  sif(true, I0, I1) -> I1 + sum(I0, I1 + 1) 
  sum(I2, I3) -> sif(I2 >= I3, I2, I3) 

This problem is converted using chaining, where edges between chained DPs are removed.

DP problem for innermost termination.
P =
  sif#(true, I0, I1) -> sum#(I0, I1 + 1) 
  sum#(I2, I3) -> sif#(I2 >= I3, I2, I3) 
  sum#(I2, I3) -> sum#(I2, I3 + 1) [I2 >= I3]
R = 
  sif(false, x, y) -> 0 
  sif(true, I0, I1) -> I1 + sum(I0, I1 + 1) 
  sum(I2, I3) -> sif(I2 >= I3, I2, I3) 

The dependency graph for this problem is:
  0 -> 2, 1
  1 -> 
  2 -> 2, 1
Where:
  0) sif#(true, I0, I1) -> sum#(I0, I1 + 1) 
  1) sum#(I2, I3) -> sif#(I2 >= I3, I2, I3) 
  2) sum#(I2, I3) -> sum#(I2, I3 + 1) [I2 >= I3]

We have the following SCCs.
  { 2 }

DP problem for innermost termination.
P =
  sum#(I2, I3) -> sum#(I2, I3 + 1) [I2 >= I3]
R = 
  sif(false, x, y) -> 0 
  sif(true, I0, I1) -> I1 + sum(I0, I1 + 1) 
  sum(I2, I3) -> sif(I2 >= I3, I2, I3) 

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

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

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