/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 =
  cu#(true, x) -> cu#(x < 100000, x + 1) 
R = 
  cu(true, x) -> cu(x < 100000, x + 1) 

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

DP problem for innermost termination.
P =
  cu#(true, x) -> cu_1#(x) 
  cu_1#(x) -> cu#(x < 100000, x + 1) 
  cu_1#(x) -> cu_1#(x + 1) [x < 100000]
R = 
  cu(true, x) -> cu(x < 100000, x + 1) 

The dependency graph for this problem is:
  0 -> 2, 1
  1 -> 
  2 -> 2, 1
Where:
  0) cu#(true, x) -> cu_1#(x) 
  1) cu_1#(x) -> cu#(x < 100000, x + 1) 
  2) cu_1#(x) -> cu_1#(x + 1) [x < 100000]

We have the following SCCs.
  { 2 }

DP problem for innermost termination.
P =
  cu_1#(x) -> cu_1#(x + 1) [x < 100000]
R = 
  cu(true, x) -> cu(x < 100000, x + 1) 

We use the reverse value criterion with the projection function NU:
  NU[cu_1#(z1)] = 100000 + -1 * z1

This gives the following inequalities:
  x < 100000  ==>  100000 + -1 * x > 100000 + -1 * (x + 1)  with  100000 + -1 * x >= 0

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