/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 =
  f#(true, x) -> f#(1000 >= x, x + 1) 
R = 
  f(true, x) -> f(1000 >= x, x + 1) 

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

DP problem for innermost termination.
P =
  f#(true, x) -> f_1#(x) 
  f_1#(x) -> f#(1000 >= x, x + 1) 
  f_1#(x) -> f_1#(x + 1) [1000 >= x]
R = 
  f(true, x) -> f(1000 >= x, x + 1) 

The dependency graph for this problem is:
  0 -> 2, 1
  1 -> 
  2 -> 2, 1
Where:
  0) f#(true, x) -> f_1#(x) 
  1) f_1#(x) -> f#(1000 >= x, x + 1) 
  2) f_1#(x) -> f_1#(x + 1) [1000 >= x]

We have the following SCCs.
  { 2 }

DP problem for innermost termination.
P =
  f_1#(x) -> f_1#(x + 1) [1000 >= x]
R = 
  f(true, x) -> f(1000 >= x, x + 1) 

We use the reverse value criterion with the projection function NU:
  NU[f_1#(z1)] = 1000 + -1 * z1

This gives the following inequalities:
  1000 >= x  ==>  1000 + -1 * x > 1000 + -1 * (x + 1)  with  1000 + -1 * x >= 0

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