/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 =
  f#(true, x, y) -> f#(x > y, x + 1, y + 2) 
R = 
  f(true, x, y) -> f(x > y, x + 1, y + 2) 

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

DP problem for innermost termination.
P =
  f#(true, x, y) -> f_1#(x, y) 
  f_1#(x, y) -> f#(x > y, x + 1, y + 2) 
  f_1#(x, y) -> f_1#(x + 1, y + 2) [x > y]
R = 
  f(true, x, y) -> f(x > y, x + 1, y + 2) 

The dependency graph for this problem is:
  0 -> 2, 1
  1 -> 
  2 -> 2, 1
Where:
  0) f#(true, x, y) -> f_1#(x, y) 
  1) f_1#(x, y) -> f#(x > y, x + 1, y + 2) 
  2) f_1#(x, y) -> f_1#(x + 1, y + 2) [x > y]

We have the following SCCs.
  { 2 }

DP problem for innermost termination.
P =
  f_1#(x, y) -> f_1#(x + 1, y + 2) [x > y]
R = 
  f(true, x, y) -> f(x > y, x + 1, y + 2) 

We use the reverse value criterion with the projection function NU:
  NU[f_1#(z1,z2)] = z1 + -1 * z2

This gives the following inequalities:
  x > y  ==>  x + -1 * y > x + 1 + -1 * (y + 2)  with  x + -1 * y >= 0

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