/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files


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MAYBE

DP problem for innermost termination.
P =
  length#(cons(x, l)) -> length#(l) 
  cond#(false, n, B2) -> tail#(nthtail(n + 1, B2)) 
  cond#(false, n, B2) -> nthtail#(n + 1, B2) 
  nthtail#(I1, B4) -> cond#(I1 >= length(B4), I1, B4) 
  nthtail#(I1, B4) -> length#(B4) 
R = 
  length(cons(x, l)) -> 1 + length(l) 
  length(nil) -> 0 
  tail(cons(B0, B1)) -> B1 
  tail(nil) -> nil 
  cond(false, n, B2) -> tail(nthtail(n + 1, B2)) 
  cond(true, I0, B3) -> B3 
  nthtail(I1, B4) -> cond(I1 >= length(B4), I1, B4) 

The dependency graph for this problem is:
  0 -> 0
  1 -> 
  2 -> 3, 4
  3 -> 1, 2
  4 -> 0
Where:
  0) length#(cons(x, l)) -> length#(l) 
  1) cond#(false, n, B2) -> tail#(nthtail(n + 1, B2)) 
  2) cond#(false, n, B2) -> nthtail#(n + 1, B2) 
  3) nthtail#(I1, B4) -> cond#(I1 >= length(B4), I1, B4) 
  4) nthtail#(I1, B4) -> length#(B4) 

We have the following SCCs.
  { 2, 3 }
  { 0 }

DP problem for innermost termination.
P =
  length#(cons(x, l)) -> length#(l) 
R = 
  length(cons(x, l)) -> 1 + length(l) 
  length(nil) -> 0 
  tail(cons(B0, B1)) -> B1 
  tail(nil) -> nil 
  cond(false, n, B2) -> tail(nthtail(n + 1, B2)) 
  cond(true, I0, B3) -> B3 
  nthtail(I1, B4) -> cond(I1 >= length(B4), I1, B4) 

We use the subterm criterion with the projection function NU:
  NU[length#(x0)] = x0

This gives the following inequalities:
  cons(x, l) |> l

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

DP problem for innermost termination.
P =
  cond#(false, n, B2) -> nthtail#(n + 1, B2) 
  nthtail#(I1, B4) -> cond#(I1 >= length(B4), I1, B4) 
R = 
  length(cons(x, l)) -> 1 + length(l) 
  length(nil) -> 0 
  tail(cons(B0, B1)) -> B1 
  tail(nil) -> nil 
  cond(false, n, B2) -> tail(nthtail(n + 1, B2)) 
  cond(true, I0, B3) -> B3 
  nthtail(I1, B4) -> cond(I1 >= length(B4), I1, B4)