/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 =
  ft#(true, x, y) -> ft#(y >= x, x + 1, y) 
  f#(true, I0) -> ft#(true, I0, I1) 
R = 
  ft(true, x, y) -> ft(y >= x, x + 1, y) 
  f(true, I0) -> ft(true, I0, I1) 

This problem is converted using chaining, where edges between chained DPs are removed.

DP problem for innermost termination.
P =
  ft#(true, x, y) -> ft_1#(x, y) 
  f#(true, I0) -> ft#(true, I0, I1) 
  ft_1#(x, y) -> ft#(y >= x, x + 1, y) 
  ft_1#(x, y) -> ft_1#(x + 1, y) [y >= x]
  f#(true, I0) -> ft_1#(I0, I1) [true]
R = 
  ft(true, x, y) -> ft(y >= x, x + 1, y) 
  f(true, I0) -> ft(true, I0, I1) 

The dependency graph for this problem is:
  0 -> 3, 2
  1 -> 
  2 -> 
  3 -> 3, 2
  4 -> 2, 3
Where:
  0) ft#(true, x, y) -> ft_1#(x, y) 
  1) f#(true, I0) -> ft#(true, I0, I1) 
  2) ft_1#(x, y) -> ft#(y >= x, x + 1, y) 
  3) ft_1#(x, y) -> ft_1#(x + 1, y) [y >= x]
  4) f#(true, I0) -> ft_1#(I0, I1) [true]

We have the following SCCs.
  { 3 }

DP problem for innermost termination.
P =
  ft_1#(x, y) -> ft_1#(x + 1, y) [y >= x]
R = 
  ft(true, x, y) -> ft(y >= x, x + 1, y) 
  f(true, I0) -> ft(true, I0, I1) 

We use the reverse value criterion with the projection function NU:
  NU[ft_1#(z1,z2)] = z2 + -1 * z1

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

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