NO

### Pre-processing the ITS problem ###

Initial linear ITS problem
   Start location: l6
      0: l0 -> l1 : L^0'=L^post_1, n^0'=n^post_1, o^0'=o^post_1, [ 1+o^0<=n^0 && L^0==L^post_1 && n^0==n^post_1 && o^0==o^post_1 ], cost: 1
      1: l0 -> l2 : L^0'=L^post_2, n^0'=n^post_2, o^0'=o^post_2, [ n^0<=o^0 && L^0==L^post_2 && n^0==n^post_2 && o^0==o^post_2 ], cost: 1
      4: l1 -> l4 : L^0'=L^post_5, n^0'=n^post_5, o^0'=o^post_5, [ L^0<=1 && 1<=L^0 && L^0==L^post_5 && n^0==n^post_5 && o^0==o^post_5 ], cost: 1
      5: l1 -> l3 : L^0'=L^post_6, n^0'=n^post_6, o^0'=o^post_6, [ L^post_6==1 && o^post_6==n^0 && n^0==n^post_6 ], cost: 1
      2: l3 -> l0 : L^0'=L^post_3, n^0'=n^post_3, o^0'=o^post_3, [ n^0<=n^0 && L^0==L^post_3 && n^0==n^post_3 && o^0==o^post_3 ], cost: 1
      3: l3 -> l0 : L^0'=L^post_4, n^0'=n^post_4, o^0'=o^post_4, [ L^post_4==0 && n^post_4==1+n^0 && o^0==o^post_4 ], cost: 1
      6: l5 -> l1 : L^0'=L^post_7, n^0'=n^post_7, o^0'=o^post_7, [ L^post_7==0 && n^0==n^post_7 && o^0==o^post_7 ], cost: 1
      7: l6 -> l5 : L^0'=L^post_8, n^0'=n^post_8, o^0'=o^post_8, [ L^0==L^post_8 && n^0==n^post_8 && o^0==o^post_8 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      7: l6 -> l5 : L^0'=L^post_8, n^0'=n^post_8, o^0'=o^post_8, [ L^0==L^post_8 && n^0==n^post_8 && o^0==o^post_8 ], cost: 1

Removed unreachable and leaf rules:
   Start location: l6
      0: l0 -> l1 : L^0'=L^post_1, n^0'=n^post_1, o^0'=o^post_1, [ 1+o^0<=n^0 && L^0==L^post_1 && n^0==n^post_1 && o^0==o^post_1 ], cost: 1
      5: l1 -> l3 : L^0'=L^post_6, n^0'=n^post_6, o^0'=o^post_6, [ L^post_6==1 && o^post_6==n^0 && n^0==n^post_6 ], cost: 1
      2: l3 -> l0 : L^0'=L^post_3, n^0'=n^post_3, o^0'=o^post_3, [ n^0<=n^0 && L^0==L^post_3 && n^0==n^post_3 && o^0==o^post_3 ], cost: 1
      3: l3 -> l0 : L^0'=L^post_4, n^0'=n^post_4, o^0'=o^post_4, [ L^post_4==0 && n^post_4==1+n^0 && o^0==o^post_4 ], cost: 1
      6: l5 -> l1 : L^0'=L^post_7, n^0'=n^post_7, o^0'=o^post_7, [ L^post_7==0 && n^0==n^post_7 && o^0==o^post_7 ], cost: 1
      7: l6 -> l5 : L^0'=L^post_8, n^0'=n^post_8, o^0'=o^post_8, [ L^0==L^post_8 && n^0==n^post_8 && o^0==o^post_8 ], cost: 1

Simplified all rules, resulting in:
   Start location: l6
      0: l0 -> l1 : [ 1+o^0<=n^0 ], cost: 1
      5: l1 -> l3 : L^0'=1, o^0'=n^0, [], cost: 1
      2: l3 -> l0 : [], cost: 1
      3: l3 -> l0 : L^0'=0, n^0'=1+n^0, [], cost: 1
      6: l5 -> l1 : L^0'=0, [], cost: 1
      7: l6 -> l5 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l6
      0: l0 -> l1 : [ 1+o^0<=n^0 ], cost: 1
      5: l1 -> l3 : L^0'=1, o^0'=n^0, [], cost: 1
      2: l3 -> l0 : [], cost: 1
      3: l3 -> l0 : L^0'=0, n^0'=1+n^0, [], cost: 1
      8: l6 -> l1 : L^0'=0, [], cost: 2

Eliminated locations (on tree-shaped paths):
   Start location: l6
      0: l0 -> l1 : [ 1+o^0<=n^0 ], cost: 1
      9: l1 -> l0 : L^0'=1, o^0'=n^0, [], cost: 2
     10: l1 -> l0 : L^0'=0, n^0'=1+n^0, o^0'=n^0, [], cost: 2
      8: l6 -> l1 : L^0'=0, [], cost: 2

Eliminated locations (on tree-shaped paths):
   Start location: l6
     11: l1 -> l1 : L^0'=0, n^0'=1+n^0, o^0'=n^0, [], cost: 3
      8: l6 -> l1 : L^0'=0, [], cost: 2

Accelerating simple loops of location 1.
   Accelerating the following rules:
     11: l1 -> l1 : L^0'=0, n^0'=1+n^0, o^0'=n^0, [], cost: 3

   Accelerated rule 11 with non-termination, yielding the new rule 12.
[accelerate] Nesting with 0 inner and 0 outer candidates
   Removing the simple loops: 11.

Accelerated all simple loops using metering functions (where possible):
   Start location: l6
     12: l1 -> [7] : [], cost: NONTERM
      8: l6 -> l1 : L^0'=0, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l6
      8: l6 -> l1 : L^0'=0, [], cost: 2
     13: l6 -> [7] : [], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l6
     13: l6 -> [7] : [], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l6
     13: l6 -> [7] : [], cost: NONTERM

Computing asymptotic complexity for rule 13
   Guard is satisfiable, yielding nontermination
   Resulting cost NONTERM has complexity: Nonterm

Found new complexity Nonterm.

Obtained the following overall complexity (w.r.t. the length of the input n):
   Complexity:  Nonterm
   Cpx degree:  Nonterm
   Solved cost: NONTERM
   Rule cost:   NONTERM
   Rule guard:  []

NO