/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 =
  if#(true, x, y) -> div#(x - y, y) 
  div#(I0, I1) -> if#(I0 >= I1 && I1 > 0, I0, I1) 
R = 
  if(true, x, y) -> div(x - y, y) + 1 
  div(I0, I1) -> if(I0 >= I1 && I1 > 0, I0, I1) 

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

DP problem for innermost termination.
P =
  if#(true, x, y) -> div#(x - y, y) 
  div#(I0, I1) -> if#(I0 >= I1 && I1 > 0, I0, I1) 
  div#(I0, I1) -> div#(I0 - I1, I1) [I0 >= I1 && I1 > 0]
R = 
  if(true, x, y) -> div(x - y, y) + 1 
  div(I0, I1) -> if(I0 >= I1 && I1 > 0, I0, I1) 

The dependency graph for this problem is:
  0 -> 2, 1
  1 -> 
  2 -> 2, 1
Where:
  0) if#(true, x, y) -> div#(x - y, y) 
  1) div#(I0, I1) -> if#(I0 >= I1 && I1 > 0, I0, I1) 
  2) div#(I0, I1) -> div#(I0 - I1, I1) [I0 >= I1 && I1 > 0]

We have the following SCCs.
  { 2 }

DP problem for innermost termination.
P =
  div#(I0, I1) -> div#(I0 - I1, I1) [I0 >= I1 && I1 > 0]
R = 
  if(true, x, y) -> div(x - y, y) + 1 
  div(I0, I1) -> if(I0 >= I1 && I1 > 0, I0, I1) 

We use the reverse value criterion with the projection function NU:
  NU[div#(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.