/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 =
  cd#(true, x) -> cd#(x > 0, x - 1) 
R = 
  cd(true, x) -> cd(x > 0, x - 1) 

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

DP problem for innermost termination.
P =
  cd#(true, x) -> cd_1#(x) 
  cd_1#(x) -> cd#(x > 0, x - 1) 
  cd_1#(x) -> cd_1#(x - 1) [x > 0]
R = 
  cd(true, x) -> cd(x > 0, x - 1) 

The dependency graph for this problem is:
  0 -> 2, 1
  1 -> 
  2 -> 2, 1
Where:
  0) cd#(true, x) -> cd_1#(x) 
  1) cd_1#(x) -> cd#(x > 0, x - 1) 
  2) cd_1#(x) -> cd_1#(x - 1) [x > 0]

We have the following SCCs.
  { 2 }

DP problem for innermost termination.
P =
  cd_1#(x) -> cd_1#(x - 1) [x > 0]
R = 
  cd(true, x) -> cd(x > 0, x - 1) 

We use the reverse value criterion with the projection function NU:
  NU[cd_1#(z1)] = z1

This gives the following inequalities:
  x > 0  ==>  x > x - 1  with  x >= 0

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