/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#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an]
  eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0]
  eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5]
  eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10]
R = 
  eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an]
  eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0]
  eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5]
  eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10]

The dependency graph for this problem is:
  0 -> 0
  1 -> 1
  2 -> 0, 2, 3
  3 -> 1, 2, 3
Where:
  0) eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an]
  1) eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0]
  2) eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5]
  3) eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10]

We have the following SCCs.
  { 2, 3 }
  { 0 }
  { 1 }

DP problem for innermost termination.
P =
  eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0]
R = 
  eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an]
  eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0]
  eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5]
  eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10]

We use the reverse value criterion with the projection function NU:
  NU[eval#(z1,z2,z3,z4,z5)] = z1 + -1 * z4

This gives the following inequalities:
  I4 >= I1 && I3 < I0  ==>  I0 + -1 * I3 > I0 + -1 * (I3 + 1)  with  I0 + -1 * I3 >= 0

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

DP problem for innermost termination.
P =
  eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an]
R = 
  eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an]
  eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0]
  eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5]
  eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10]

We use the reverse value criterion with the projection function NU:
  NU[eval#(z1,z2,z3,z4,z5)] = z2 + -1 * z5

This gives the following inequalities:
  j < bn && i >= an  ==>  bn + -1 * j > bn + -1 * (j + 1)  with  bn + -1 * j >= 0

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

DP problem for innermost termination.
P =
  eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5]
  eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10]
R = 
  eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an]
  eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0]
  eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5]
  eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10]

We use the reverse value criterion with the projection function NU:
  NU[eval#(z1,z2,z3,z4,z5)] = z1 + -1 * z4

This gives the following inequalities:
  I9 < I6 && I8 < I5  ==>  I5 + -1 * I8 > I5 + -1 * (I8 + 1)  with  I5 + -1 * I8 >= 0
  I14 < I11 && I13 < I10  ==>  I10 + -1 * I13 >= I10 + -1 * I13

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10]
R = 
  eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an]
  eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0]
  eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5]
  eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10]

We use the reverse value criterion with the projection function NU:
  NU[eval#(z1,z2,z3,z4,z5)] = z2 + -1 * z5

This gives the following inequalities:
  I14 < I11 && I13 < I10  ==>  I11 + -1 * I14 > I11 + -1 * (I14 + 1)  with  I11 + -1 * I14 >= 0

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