NO

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

Initial linear ITS problem
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ 5==arg1P_1 && arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && 3==arg2P_2 && arg3==arg3P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && 0==arg3P_3 ], cost: 1
      4: f4 -> f5 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg3<10 && arg1==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      6: f4 -> f7 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg3>=10 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg3P_4==arg3 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      5: f6 -> f4 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [], cost: 1

Removed unreachable and leaf rules:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ 5==arg1P_1 && arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && 3==arg2P_2 && arg3==arg3P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && 0==arg3P_3 ], cost: 1
      4: f4 -> f5 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg3<10 && arg1==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg3P_4==arg3 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      5: f6 -> f4 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg1'=5, [], cost: 1
      1: f2 -> f3 : arg2'=3, [], cost: 1
      2: f3 -> f4 : arg3'=0, [], cost: 1
      4: f4 -> f5 : [ arg3<10 ], cost: 1
      3: f5 -> f6 : [], cost: 1
      5: f6 -> f4 : [], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     12: f4 -> f4 : [ arg3<10 ], cost: 3
     10: __init -> f4 : arg1'=5, arg2'=3, arg3'=0, [], cost: 4

Accelerating simple loops of location 3.
   Accelerating the following rules:
     12: f4 -> f4 : [ arg3<10 ], cost: 3

   Accelerated rule 12 with non-termination, yielding the new rule 13.
[accelerate] Nesting with 0 inner and 0 outer candidates
   Removing the simple loops: 12.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     13: f4 -> [8] : [ arg3<10 ], cost: NONTERM
     10: __init -> f4 : arg1'=5, arg2'=3, arg3'=0, [], cost: 4

Chained accelerated rules (with incoming rules):
   Start location: __init
     10: __init -> f4 : arg1'=5, arg2'=3, arg3'=0, [], cost: 4
     14: __init -> [8] : [], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     14: __init -> [8] : [], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     14: __init -> [8] : [], cost: NONTERM

Computing asymptotic complexity for rule 14
   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