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

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

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

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

We use the reverse value criterion with the projection function NU:
  NU[eval#(z1,z2)] = z1 + z2 + -1 * 0

This gives the following inequalities:
  I0 + I1 > 0 && I1 >= I0 && I1 > I0  ==>  I0 + I1 + -1 * 0 > I0 + (I1 - 1) + -1 * 0  with  I0 + I1 + -1 * 0 >= 0
  I2 + I3 > 0 && I3 >= I2 && I2 = I3  ==>  I2 + I3 + -1 * 0 > I2 - 1 + I3 + -1 * 0  with  I2 + I3 + -1 * 0 >= 0

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

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

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

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

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