/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_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x]
  eval_2#(I0, I1, I2) -> eval_1#(I0, I1, I2 + 1) [I0 > I2]
  eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5]
  eval_1#(I6, I7, I8) -> eval_2#(I6, I7, I8) [I6 > I7]
R = 
  eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x]
  eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2]
  eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5]
  eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7]

We use the reverse value criterion with the projection function NU:
  NU[eval_1#(z1,z2,z3)] = z1 + -1 * z3
  NU[eval_2#(z1,z2,z3)] = z1 + -1 * z3

This gives the following inequalities:
  z >= x  ==>  x + -1 * z >= x - 1 + -1 * z
  I0 > I2  ==>  I0 + -1 * I2 > I0 + -1 * (I2 + 1)  with  I0 + -1 * I2 >= 0
  I3 > I5  ==>  I3 + -1 * I5 >= I3 + -1 * I5
  I6 > I7  ==>  I6 + -1 * I8 >= I6 + -1 * I8

We remove all the strictly oriented dependency pairs.

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

We use the reverse value criterion with the projection function NU:
  NU[eval_1#(z1,z2,z3)] = z1 + -1 * z2
  NU[eval_2#(z1,z2,z3)] = z1 - 1 + -1 * z2

This gives the following inequalities:
  z >= x  ==>  x - 1 + -1 * y >= x - 1 + -1 * y
  I3 > I5  ==>  I3 - 1 + -1 * I4 >= I3 + -1 * (I4 + 1)
  I6 > I7  ==>  I6 + -1 * I7 > I6 - 1 + -1 * I7  with  I6 + -1 * I7 >= 0

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  0 -> 
  2 -> 
Where:
  0) eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x]
  2) eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5]

We have the following SCCs.