/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, y, z) -> eval#(x, y, z) [y > z && z >= x && z >= y]
  eval#(I0, I1, I2) -> eval#(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
  eval#(I3, I4, I5) -> eval#(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
  eval#(I6, I7, I8) -> eval#(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
  eval#(I9, I10, I11) -> eval#(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
  eval#(I12, I13, I14) -> eval#(I12 - 1, I13, I14) [I12 > I14]
R = 
  eval(x, y, z) -> eval(x, y, z) [y > z && z >= x && z >= y]
  eval(I0, I1, I2) -> eval(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
  eval(I3, I4, I5) -> eval(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
  eval(I6, I7, I8) -> eval(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
  eval(I9, I10, I11) -> eval(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
  eval(I12, I13, I14) -> eval(I12 - 1, I13, I14) [I12 > I14]

The dependency graph for this problem is:
  0 -> 
  1 -> 
  2 -> 2
  3 -> 
  4 -> 2, 4, 5
  5 -> 2, 4, 5
Where:
  0) eval#(x, y, z) -> eval#(x, y, z) [y > z && z >= x && z >= y]
  1) eval#(I0, I1, I2) -> eval#(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
  2) eval#(I3, I4, I5) -> eval#(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
  3) eval#(I6, I7, I8) -> eval#(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
  4) eval#(I9, I10, I11) -> eval#(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
  5) eval#(I12, I13, I14) -> eval#(I12 - 1, I13, I14) [I12 > I14]

We have the following SCCs.
  { 4, 5 }
  { 2 }

DP problem for innermost termination.
P =
  eval#(I3, I4, I5) -> eval#(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
R = 
  eval(x, y, z) -> eval(x, y, z) [y > z && z >= x && z >= y]
  eval(I0, I1, I2) -> eval(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
  eval(I3, I4, I5) -> eval(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
  eval(I6, I7, I8) -> eval(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
  eval(I9, I10, I11) -> eval(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
  eval(I12, I13, I14) -> eval(I12 - 1, I13, I14) [I12 > I14]

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

This gives the following inequalities:
  I4 > I5 && I5 >= I3  ==>  I4 + -1 * I5 > I4 - 1 + -1 * I5  with  I4 + -1 * I5 >= 0

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

DP problem for innermost termination.
P =
  eval#(I9, I10, I11) -> eval#(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
  eval#(I12, I13, I14) -> eval#(I12 - 1, I13, I14) [I12 > I14]
R = 
  eval(x, y, z) -> eval(x, y, z) [y > z && z >= x && z >= y]
  eval(I0, I1, I2) -> eval(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
  eval(I3, I4, I5) -> eval(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
  eval(I6, I7, I8) -> eval(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
  eval(I9, I10, I11) -> eval(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
  eval(I12, I13, I14) -> eval(I12 - 1, I13, I14) [I12 > I14]

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

This gives the following inequalities:
  I10 > I11 && I9 > I11  ==>  I9 + -1 * I11 > I9 - 1 + -1 * I11  with  I9 + -1 * I11 >= 0
  I12 > I14  ==>  I12 + -1 * I14 > I12 - 1 + -1 * I14  with  I12 + -1 * I14 >= 0

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