/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 =
  cond#(true, I0, I1, I2) -> d#(I0, I1 + 1, I1 + 1 + I2) 
  dNat#(true, I3, I4, I5) -> cond#(I3 >= I5, I3, I4 - 1, I5) 
  d#(I6, I7, I8) -> dNat#(I6 >= 0 && I7 >= 1 && I8 >= 0, I6, I7, I8) 
  divNat#(true, I9, I10) -> d#(I9, I10, 0) 
  div#(I11, I12) -> divNat#(I11 >= 0 && I12 >= 1, I11, I12) 
R = 
  cond(false, x, y, z) -> 0 
  cond(true, I0, I1, I2) -> 1 + d(I0, I1 + 1, I1 + 1 + I2) 
  dNat(true, I3, I4, I5) -> cond(I3 >= I5, I3, I4 - 1, I5) 
  d(I6, I7, I8) -> dNat(I6 >= 0 && I7 >= 1 && I8 >= 0, I6, I7, I8) 
  divNat(true, I9, I10) -> d(I9, I10, 0) 
  div(I11, I12) -> divNat(I11 >= 0 && I12 >= 1, I11, I12) 

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

DP problem for innermost termination.
P =
  cond#(true, I0, I1, I2) -> d#(I0, I1 + 1, I1 + 1 + I2) 
  dNat#(true, I3, I4, I5) -> cond#(I3 >= I5, I3, I4 - 1, I5) 
  d#(I6, I7, I8) -> dNat#(I6 >= 0 && I7 >= 1 && I8 >= 0, I6, I7, I8) 
  divNat#(true, I9, I10) -> d#(I9, I10, 0) 
  div#(I11, I12) -> divNat#(I11 >= 0 && I12 >= 1, I11, I12) 
  div#(I11, I12) -> d#(I11, I12, 0) [I11 >= 0 && I12 >= 1]
  d#(I6, I7, I8) -> cond#(I6 >= I8, I6, I7 - 1, I8) [I6 >= 0 && I7 >= 1 && I8 >= 0]
  d#(I6, I7, I8) -> d#(I6, I7 - 1 + 1, I7 - 1 + 1 + I8) [I6 >= 0 && I7 >= 1 && I8 >= 0, I6 >= I8]
  dNat#(true, I3, I4, I5) -> d#(I3, I4 - 1 + 1, I4 - 1 + 1 + I5) [I3 >= I5]
R = 
  cond(false, x, y, z) -> 0 
  cond(true, I0, I1, I2) -> 1 + d(I0, I1 + 1, I1 + 1 + I2) 
  dNat(true, I3, I4, I5) -> cond(I3 >= I5, I3, I4 - 1, I5) 
  d(I6, I7, I8) -> dNat(I6 >= 0 && I7 >= 1 && I8 >= 0, I6, I7, I8) 
  divNat(true, I9, I10) -> d(I9, I10, 0) 
  div(I11, I12) -> divNat(I11 >= 0 && I12 >= 1, I11, I12) 

The dependency graph for this problem is:
  0 -> 7, 6, 2
  1 -> 
  2 -> 
  3 -> 7, 6, 2
  4 -> 
  5 -> 7, 6, 2
  6 -> 
  7 -> 7, 2, 6
  8 -> 2, 6, 7
Where:
  0) cond#(true, I0, I1, I2) -> d#(I0, I1 + 1, I1 + 1 + I2) 
  1) dNat#(true, I3, I4, I5) -> cond#(I3 >= I5, I3, I4 - 1, I5) 
  2) d#(I6, I7, I8) -> dNat#(I6 >= 0 && I7 >= 1 && I8 >= 0, I6, I7, I8) 
  3) divNat#(true, I9, I10) -> d#(I9, I10, 0) 
  4) div#(I11, I12) -> divNat#(I11 >= 0 && I12 >= 1, I11, I12) 
  5) div#(I11, I12) -> d#(I11, I12, 0) [I11 >= 0 && I12 >= 1]
  6) d#(I6, I7, I8) -> cond#(I6 >= I8, I6, I7 - 1, I8) [I6 >= 0 && I7 >= 1 && I8 >= 0]
  7) d#(I6, I7, I8) -> d#(I6, I7 - 1 + 1, I7 - 1 + 1 + I8) [I6 >= 0 && I7 >= 1 && I8 >= 0, I6 >= I8]
  8) dNat#(true, I3, I4, I5) -> d#(I3, I4 - 1 + 1, I4 - 1 + 1 + I5) [I3 >= I5]

We have the following SCCs.
  { 7 }

DP problem for innermost termination.
P =
  d#(I6, I7, I8) -> d#(I6, I7 - 1 + 1, I7 - 1 + 1 + I8) [I6 >= 0 && I7 >= 1 && I8 >= 0, I6 >= I8]
R = 
  cond(false, x, y, z) -> 0 
  cond(true, I0, I1, I2) -> 1 + d(I0, I1 + 1, I1 + 1 + I2) 
  dNat(true, I3, I4, I5) -> cond(I3 >= I5, I3, I4 - 1, I5) 
  d(I6, I7, I8) -> dNat(I6 >= 0 && I7 >= 1 && I8 >= 0, I6, I7, I8) 
  divNat(true, I9, I10) -> d(I9, I10, 0) 
  div(I11, I12) -> divNat(I11 >= 0 && I12 >= 1, I11, I12) 

We use the reverse value criterion with the projection function NU:
  NU[d#(z1,z2,z3)] = z1 + -1 * z3

This gives the following inequalities:
  I6 >= 0 && I7 >= 1 && I8 >= 0/\I6 >= I8  ==>  I6 + -1 * I8 > I6 + -1 * (I7 - 1 + 1 + I8)  with  I6 + -1 * I8 >= 0

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