NO

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

Initial linear ITS problem
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1
      7: f2 -> f3 : arg1'=arg1P_8, [ arg1==arg1P_8 ], cost: 1
      1: f4 -> f7 : arg1'=arg1P_2, [ arg1P_2==-1+arg1 ], cost: 1
      5: f7 -> f6 : arg1'=arg1P_6, [ arg1==arg1P_6 ], cost: 1
      2: f5 -> f8 : arg1'=arg1P_3, [ arg1P_3==1+arg1 ], cost: 1
      6: f8 -> f6 : arg1'=arg1P_7, [ arg1==arg1P_7 ], cost: 1
      3: f3 -> f4 : arg1'=arg1P_4, [ arg1<=0 && arg1==arg1P_4 ], cost: 1
      4: f3 -> f5 : arg1'=arg1P_5, [ arg1>0 && arg1==arg1P_5 ], cost: 1
      8: f6 -> f2 : arg1'=arg1P_9, [ arg1==arg1P_9 ], cost: 1
      9: __init -> f1 : arg1'=arg1P_10, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      9: __init -> f1 : arg1'=arg1P_10, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1
      7: f2 -> f3 : [], cost: 1
      1: f4 -> f7 : arg1'=-1+arg1, [], cost: 1
      5: f7 -> f6 : [], cost: 1
      2: f5 -> f8 : arg1'=1+arg1, [], cost: 1
      6: f8 -> f6 : [], cost: 1
      3: f3 -> f4 : [ arg1<=0 ], cost: 1
      4: f3 -> f5 : [ arg1>0 ], cost: 1
      8: f6 -> f2 : [], cost: 1
      9: __init -> f1 : arg1'=arg1P_10, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
      7: f2 -> f3 : [], cost: 1
     13: f3 -> f6 : arg1'=-1+arg1, [ arg1<=0 ], cost: 3
     14: f3 -> f6 : arg1'=1+arg1, [ arg1>0 ], cost: 3
      8: f6 -> f2 : [], cost: 1
     10: __init -> f2 : arg1'=arg1P_1, [], cost: 2

Eliminated locations (on tree-shaped paths):
   Start location: __init
     15: f2 -> f6 : arg1'=-1+arg1, [ arg1<=0 ], cost: 4
     16: f2 -> f6 : arg1'=1+arg1, [ arg1>0 ], cost: 4
      8: f6 -> f2 : [], cost: 1
     10: __init -> f2 : arg1'=arg1P_1, [], cost: 2

Eliminated locations (on tree-shaped paths):
   Start location: __init
     17: f2 -> f2 : arg1'=-1+arg1, [ arg1<=0 ], cost: 5
     18: f2 -> f2 : arg1'=1+arg1, [ arg1>0 ], cost: 5
     10: __init -> f2 : arg1'=arg1P_1, [], cost: 2

Accelerating simple loops of location 1.
   Accelerating the following rules:
     17: f2 -> f2 : arg1'=-1+arg1, [ arg1<=0 ], cost: 5
     18: f2 -> f2 : arg1'=1+arg1, [ arg1>0 ], cost: 5

   Accelerated rule 17 with non-termination, yielding the new rule 19.
   Accelerated rule 18 with non-termination, yielding the new rule 20.
[accelerate] Nesting with 0 inner and 0 outer candidates
   Removing the simple loops: 17 18.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     19: f2 -> [9] : [ arg1<=0 ], cost: NONTERM
     20: f2 -> [9] : [ arg1>0 ], cost: NONTERM
     10: __init -> f2 : arg1'=arg1P_1, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: __init
     10: __init -> f2 : arg1'=arg1P_1, [], cost: 2
     21: __init -> [9] : [], cost: NONTERM
     22: __init -> [9] : [], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     21: __init -> [9] : [], cost: NONTERM
     22: __init -> [9] : [], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     22: __init -> [9] : [], cost: NONTERM

Computing asymptotic complexity for rule 22
   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