NO

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

Initial linear ITS problem
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ 5==arg1P_1 && arg2==arg2P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 ], cost: 1
      3: f3 -> f4 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2>=0 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      5: f3 -> f6 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg2<0 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==arg2+arg1 && arg1==arg1P_3 ], cost: 1
      4: f5 -> f3 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      6: __init -> f1 : arg1'=arg1P_7, arg2'=arg2P_7, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      6: __init -> f1 : arg1'=arg1P_7, arg2'=arg2P_7, [], cost: 1

Removed unreachable and leaf rules:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ 5==arg1P_1 && arg2==arg2P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 ], cost: 1
      3: f3 -> f4 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2>=0 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==arg2+arg1 && arg1==arg1P_3 ], cost: 1
      4: f5 -> f3 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      6: __init -> f1 : arg1'=arg1P_7, arg2'=arg2P_7, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg1'=5, [], cost: 1
      1: f2 -> f3 : arg2'=arg2P_2, [], cost: 1
      3: f3 -> f4 : [ arg2>=0 ], cost: 1
      2: f4 -> f5 : arg2'=arg2+arg1, [], cost: 1
      4: f5 -> f3 : [], cost: 1
      6: __init -> f1 : arg1'=arg1P_7, arg2'=arg2P_7, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     10: f3 -> f3 : arg2'=arg2+arg1, [ arg2>=0 ], cost: 3
      8: __init -> f3 : arg1'=5, arg2'=arg2P_2, [], cost: 3

Accelerating simple loops of location 2.
   Accelerating the following rules:
     10: f3 -> f3 : arg2'=arg2+arg1, [ arg2>=0 ], cost: 3

[test] deduced invariant -arg1<=0
   Accelerated rule 10 with non-termination, yielding the new rule 11.
   Accelerated rule 10 with non-termination, yielding the new rule 12.
   Accelerated rule 10 with backward acceleration, yielding the new rule 13.
[accelerate] Nesting with 0 inner and 1 outer candidates
   Also removing duplicate rules: 12.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     10: f3 -> f3 : arg2'=arg2+arg1, [ arg2>=0 ], cost: 3
     11: f3 -> [7] : [ arg2==0 && arg1==0 ], cost: NONTERM
     13: f3 -> [7] : [ arg2>=0 && -arg1<=0 ], cost: NONTERM
      8: __init -> f3 : arg1'=5, arg2'=arg2P_2, [], cost: 3

Chained accelerated rules (with incoming rules):
   Start location: __init
      8: __init -> f3 : arg1'=5, arg2'=arg2P_2, [], cost: 3
     14: __init -> f3 : arg1'=5, arg2'=5+arg2P_2, [ arg2P_2>=0 ], cost: 6
     15: __init -> [7] : [], cost: NONTERM

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

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     15: __init -> [7] : [], cost: NONTERM

Computing asymptotic complexity for rule 15
   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