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

The dependency graph for this problem is:
  0 -> 0, 3
  1 -> 
  2 -> 
  3 -> 0, 3
Where:
  0) eval#(x, y) -> eval#(x, y - x) [y > x && x > 0 && y > 0 && y >= x]
  1) eval#(I0, I1) -> eval#(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0]
  2) eval#(I2, I3) -> eval#(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3]
  3) eval#(I4, I5) -> eval#(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0]

We have the following SCCs.
  { 0, 3 }

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

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

This gives the following inequalities:
  y > x && x > 0 && y > 0 && y >= x  ==>  x >= x
  I4 > I5 && I4 > 0 && I5 > 0  ==>  I4 > I4 - I5  with  I4 >= 0

We remove all the strictly oriented dependency pairs.

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

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

This gives the following inequalities:
  y > x && x > 0 && y > 0 && y >= x  ==>  y + -1 * x > y - x + -1 * x  with  y + -1 * x >= 0

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