/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 =
  eval_3#(x, y) -> eval_1#(x, y) [0 >= y]
  eval_3#(I0, I1) -> eval_3#(I0, I1 - 1) [I1 > 0]
  eval_2#(I2, I3) -> eval_1#(I2, I3) [0 >= I2]
  eval_2#(I4, I5) -> eval_2#(I4 - 1, I5) [I4 > 0]
  eval_1#(I6, I7) -> eval_3#(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
  eval_1#(I8, I9) -> eval_2#(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]
R = 
  eval_3(x, y) -> eval_1(x, y) [0 >= y]
  eval_3(I0, I1) -> eval_3(I0, I1 - 1) [I1 > 0]
  eval_2(I2, I3) -> eval_1(I2, I3) [0 >= I2]
  eval_2(I4, I5) -> eval_2(I4 - 1, I5) [I4 > 0]
  eval_1(I6, I7) -> eval_3(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
  eval_1(I8, I9) -> eval_2(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]

The dependency graph for this problem is:
  0 -> 
  1 -> 0, 1
  2 -> 
  3 -> 2, 3
  4 -> 1
  5 -> 3
Where:
  0) eval_3#(x, y) -> eval_1#(x, y) [0 >= y]
  1) eval_3#(I0, I1) -> eval_3#(I0, I1 - 1) [I1 > 0]
  2) eval_2#(I2, I3) -> eval_1#(I2, I3) [0 >= I2]
  3) eval_2#(I4, I5) -> eval_2#(I4 - 1, I5) [I4 > 0]
  4) eval_1#(I6, I7) -> eval_3#(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
  5) eval_1#(I8, I9) -> eval_2#(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]

We have the following SCCs.
  { 1 }
  { 3 }

DP problem for innermost termination.
P =
  eval_2#(I4, I5) -> eval_2#(I4 - 1, I5) [I4 > 0]
R = 
  eval_3(x, y) -> eval_1(x, y) [0 >= y]
  eval_3(I0, I1) -> eval_3(I0, I1 - 1) [I1 > 0]
  eval_2(I2, I3) -> eval_1(I2, I3) [0 >= I2]
  eval_2(I4, I5) -> eval_2(I4 - 1, I5) [I4 > 0]
  eval_1(I6, I7) -> eval_3(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
  eval_1(I8, I9) -> eval_2(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]

We use the reverse value criterion with the projection function NU:
  NU[eval_2#(z1,z2)] = z1

This gives the following inequalities:
  I4 > 0  ==>  I4 > I4 - 1  with  I4 >= 0

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

DP problem for innermost termination.
P =
  eval_3#(I0, I1) -> eval_3#(I0, I1 - 1) [I1 > 0]
R = 
  eval_3(x, y) -> eval_1(x, y) [0 >= y]
  eval_3(I0, I1) -> eval_3(I0, I1 - 1) [I1 > 0]
  eval_2(I2, I3) -> eval_1(I2, I3) [0 >= I2]
  eval_2(I4, I5) -> eval_2(I4 - 1, I5) [I4 > 0]
  eval_1(I6, I7) -> eval_3(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
  eval_1(I8, I9) -> eval_2(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]

We use the reverse value criterion with the projection function NU:
  NU[eval_3#(z1,z2)] = z2 + -1 * 0

This gives the following inequalities:
  I1 > 0  ==>  I1 + -1 * 0 > I1 - 1 + -1 * 0  with  I1 + -1 * 0 >= 0

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