/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) [x + y > z && z >= 0 && 0 >= x && 0 >= y]
  eval#(I0, I1, I2) -> eval#(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0]
  eval#(I3, I4, I5) -> eval#(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0]
R = 
  eval(x, y, z) -> eval(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y]
  eval(I0, I1, I2) -> eval(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0]
  eval(I3, I4, I5) -> eval(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0]

The dependency graph for this problem is:
  0 -> 
  1 -> 1
  2 -> 1, 2
Where:
  0) eval#(x, y, z) -> eval#(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y]
  1) eval#(I0, I1, I2) -> eval#(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0]
  2) eval#(I3, I4, I5) -> eval#(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0]

We have the following SCCs.
  { 2 }
  { 1 }

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

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

This gives the following inequalities:
  I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0  ==>  I1 > I1 - 1  with  I1 >= 0

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

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

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

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

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