/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  eval#(x, y) -> eval#(x, y) [y > 0 && 0 >= x && 0 >= y]
  eval#(I0, I1) -> eval#(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
  eval#(I2, I3) -> eval#(I2, I3 - 1) [I3 > 0 && 0 >= I2]
  eval#(I4, I5) -> eval#(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
  eval#(I6, I7) -> eval#(I6 - 1, I7) [I7 > 0 && I6 > 0]
  eval#(I8, I9) -> eval#(I8 - 1, I9) [I8 > 0]
R = 
  eval(x, y) -> eval(x, y) [y > 0 && 0 >= x && 0 >= y]
  eval(I0, I1) -> eval(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
  eval(I2, I3) -> eval(I2, I3 - 1) [I3 > 0 && 0 >= I2]
  eval(I4, I5) -> eval(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
  eval(I6, I7) -> eval(I6 - 1, I7) [I7 > 0 && I6 > 0]
  eval(I8, I9) -> eval(I8 - 1, I9) [I8 > 0]

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) -> eval#(x, y) [y > 0 && 0 >= x && 0 >= y]
  1) eval#(I0, I1) -> eval#(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
  2) eval#(I2, I3) -> eval#(I2, I3 - 1) [I3 > 0 && 0 >= I2]
  3) eval#(I4, I5) -> eval#(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
  4) eval#(I6, I7) -> eval#(I6 - 1, I7) [I7 > 0 && I6 > 0]
  5) eval#(I8, I9) -> eval#(I8 - 1, I9) [I8 > 0]

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

DP problem for innermost termination.
P =
  eval#(I2, I3) -> eval#(I2, I3 - 1) [I3 > 0 && 0 >= I2]
R = 
  eval(x, y) -> eval(x, y) [y > 0 && 0 >= x && 0 >= y]
  eval(I0, I1) -> eval(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
  eval(I2, I3) -> eval(I2, I3 - 1) [I3 > 0 && 0 >= I2]
  eval(I4, I5) -> eval(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
  eval(I6, I7) -> eval(I6 - 1, I7) [I7 > 0 && I6 > 0]
  eval(I8, I9) -> eval(I8 - 1, I9) [I8 > 0]

We use the reverse value criterion with the projection function NU:
  NU[eval#(z1,z2)] = z2 + -1 * 0

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

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

DP problem for innermost termination.
P =
  eval#(I6, I7) -> eval#(I6 - 1, I7) [I7 > 0 && I6 > 0]
  eval#(I8, I9) -> eval#(I8 - 1, I9) [I8 > 0]
R = 
  eval(x, y) -> eval(x, y) [y > 0 && 0 >= x && 0 >= y]
  eval(I0, I1) -> eval(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
  eval(I2, I3) -> eval(I2, I3 - 1) [I3 > 0 && 0 >= I2]
  eval(I4, I5) -> eval(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
  eval(I6, I7) -> eval(I6 - 1, I7) [I7 > 0 && I6 > 0]
  eval(I8, I9) -> eval(I8 - 1, I9) [I8 > 0]

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

This gives the following inequalities:
  I7 > 0 && I6 > 0  ==>  I6 > I6 - 1  with  I6 >= 0
  I8 > 0  ==>  I8 > I8 - 1  with  I8 >= 0

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