/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#(x, y) -> eval#(x - 1, y) [x > 0 && not(x = 0) && x <= y]
  eval#(I0, I1) -> eval#(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1]
R = 
  eval(x, y) -> eval(x - 1, y) [x > 0 && not(x = 0) && x <= y]
  eval(I0, I1) -> eval(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1]

The dependency graph for this problem is:
  0 -> 0
  1 -> 0
Where:
  0) eval#(x, y) -> eval#(x - 1, y) [x > 0 && not(x = 0) && x <= y]
  1) eval#(I0, I1) -> eval#(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1]

We have the following SCCs.
  { 0 }

DP problem for innermost termination.
P =
  eval#(x, y) -> eval#(x - 1, y) [x > 0 && not(x = 0) && x <= y]
R = 
  eval(x, y) -> eval(x - 1, y) [x > 0 && not(x = 0) && x <= y]
  eval(I0, I1) -> eval(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1]

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

This gives the following inequalities:
  x > 0 && not(x = 0) && x <= y  ==>  x > x - 1  with  x >= 0

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