NO

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

Initial linear ITS problem
   Start location: l5
      0: l0 -> l1 : n^0'=n^post_1, x^0'=x^post_1, [ n^0==n^post_1 && x^0==x^post_1 ], cost: 1
      1: l1 -> l0 : n^0'=n^post_2, x^0'=x^post_2, [ n^0==n^post_2 && x^0==x^post_2 ], cost: 1
      2: l2 -> l0 : n^0'=n^post_3, x^0'=x^post_3, [ n^0<=0 && x^1_1==0 && x^post_3==1 && n^0==n^post_3 ], cost: 1
      3: l2 -> l3 : n^0'=n^post_4, x^0'=x^post_4, [ 1<=n^0 && n^0==n^post_4 && x^0==x^post_4 ], cost: 1
      4: l3 -> l2 : n^0'=n^post_5, x^0'=x^post_5, [ n^0==n^post_5 && x^0==x^post_5 ], cost: 1
      5: l4 -> l2 : n^0'=n^post_6, x^0'=x^post_6, [ n^post_6==n^post_6 && x^post_6==1 ], cost: 1
      6: l5 -> l4 : n^0'=n^post_7, x^0'=x^post_7, [ n^0==n^post_7 && x^0==x^post_7 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      6: l5 -> l4 : n^0'=n^post_7, x^0'=x^post_7, [ n^0==n^post_7 && x^0==x^post_7 ], cost: 1

Simplified all rules, resulting in:
   Start location: l5
      0: l0 -> l1 : [], cost: 1
      1: l1 -> l0 : [], cost: 1
      2: l2 -> l0 : x^0'=1, [ n^0<=0 ], cost: 1
      3: l2 -> l3 : [ 1<=n^0 ], cost: 1
      4: l3 -> l2 : [], cost: 1
      5: l4 -> l2 : n^0'=n^post_6, x^0'=1, [], cost: 1
      6: l5 -> l4 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l5
      9: l0 -> l0 : [], cost: 2
      2: l2 -> l0 : x^0'=1, [ n^0<=0 ], cost: 1
      8: l2 -> l2 : [ 1<=n^0 ], cost: 2
      7: l5 -> l2 : n^0'=n^post_6, x^0'=1, [], cost: 2

Accelerating simple loops of location 0.
   Accelerating the following rules:
      9: l0 -> l0 : [], cost: 2

   Accelerated rule 9 with non-termination, yielding the new rule 10.
[accelerate] Nesting with 0 inner and 0 outer candidates
   Removing the simple loops: 9.

Accelerating simple loops of location 2.
   Accelerating the following rules:
      8: l2 -> l2 : [ 1<=n^0 ], cost: 2

   Accelerated rule 8 with non-termination, yielding the new rule 11.
[accelerate] Nesting with 0 inner and 0 outer candidates
   Removing the simple loops: 8.

Accelerated all simple loops using metering functions (where possible):
   Start location: l5
     10: l0 -> [6] : [], cost: NONTERM
      2: l2 -> l0 : x^0'=1, [ n^0<=0 ], cost: 1
     11: l2 -> [7] : [ 1<=n^0 ], cost: NONTERM
      7: l5 -> l2 : n^0'=n^post_6, x^0'=1, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l5
      2: l2 -> l0 : x^0'=1, [ n^0<=0 ], cost: 1
     12: l2 -> [6] : [ n^0<=0 ], cost: NONTERM
      7: l5 -> l2 : n^0'=n^post_6, x^0'=1, [], cost: 2
     13: l5 -> [7] : [], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l5
     12: l2 -> [6] : [ n^0<=0 ], cost: NONTERM
      7: l5 -> l2 : n^0'=n^post_6, x^0'=1, [], cost: 2
     13: l5 -> [7] : [], cost: NONTERM

Eliminated locations (on linear paths):
   Start location: l5
     13: l5 -> [7] : [], cost: NONTERM
     14: l5 -> [6] : [ n^post_6<=0 ], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l5
     13: l5 -> [7] : [], cost: NONTERM
     14: l5 -> [6] : [ n^post_6<=0 ], 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