/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_1#(x, y, z) -> eval_1#(x, y, x - y) [y > x && z > y && y > 0]
  eval_1#(I0, I1, I2) -> eval_1#(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0]
  eval_0#(I3, I4, I5) -> eval_1#(I3, I4, I5) [I4 > 0]
R = 
  eval_1(x, y, z) -> eval_1(x, y, x - y) [y > x && z > y && y > 0]
  eval_1(I0, I1, I2) -> eval_1(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0]
  eval_0(I3, I4, I5) -> eval_1(I3, I4, I5) [I4 > 0]

The dependency graph for this problem is:
  0 -> 
  1 -> 0, 1
  2 -> 0, 1
Where:
  0) eval_1#(x, y, z) -> eval_1#(x, y, x - y) [y > x && z > y && y > 0]
  1) eval_1#(I0, I1, I2) -> eval_1#(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0]
  2) eval_0#(I3, I4, I5) -> eval_1#(I3, I4, I5) [I4 > 0]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  eval_1#(I0, I1, I2) -> eval_1#(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0]
R = 
  eval_1(x, y, z) -> eval_1(x, y, x - y) [y > x && z > y && y > 0]
  eval_1(I0, I1, I2) -> eval_1(I0 + I1, I1, I2) [I1 > I0 && I2 > I1 && I1 > 0]
  eval_0(I3, I4, I5) -> eval_1(I3, I4, I5) [I4 > 0]

We use the reverse value criterion with the projection function NU:
  NU[eval_1#(z1,z2,z3)] = z2 + -1 * z1

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

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