NO

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

Initial linear ITS problem
   Start location: __init
      0: f1_0_main_Load -> f53_0_loop_EQ : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1>0 && arg2>-1 && arg2==arg1P_1 ], cost: 1
      1: f53_0_loop_EQ -> f53_0_loop_EQ : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1>4 && arg1==arg1P_2 ], cost: 1
      2: f53_0_loop_EQ -> f53_0_loop_EQ : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1>0 && arg1<5 && -1+arg1==arg1P_3 ], cost: 1
      3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1_0_main_Load -> f53_0_loop_EQ : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1
      1: f53_0_loop_EQ -> f53_0_loop_EQ : arg2'=arg2P_2, [ arg1>4 ], cost: 1
      2: f53_0_loop_EQ -> f53_0_loop_EQ : arg1'=-1+arg1, arg2'=arg2P_3, [ arg1>0 && arg1<5 ], cost: 1
      3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1

### Simplification by acceleration and chaining ###

Accelerating simple loops of location 1.
   Accelerating the following rules:
      1: f53_0_loop_EQ -> f53_0_loop_EQ : arg2'=arg2P_2, [ arg1>4 ], cost: 1
      2: f53_0_loop_EQ -> f53_0_loop_EQ : arg1'=-1+arg1, arg2'=arg2P_3, [ arg1>0 && arg1<5 ], cost: 1

   Accelerated rule 1 with non-termination, yielding the new rule 4.
   Accelerated rule 2 with backward acceleration, yielding the new rule 5.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 1 2.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
      0: f1_0_main_Load -> f53_0_loop_EQ : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1
      4: f53_0_loop_EQ -> [3] : [ arg1>4 ], cost: NONTERM
      5: f53_0_loop_EQ -> f53_0_loop_EQ : arg1'=0, arg2'=arg2P_3, [ arg1<5 && arg1>=1 ], cost: arg1
      3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1

Chained accelerated rules (with incoming rules):
   Start location: __init
      0: f1_0_main_Load -> f53_0_loop_EQ : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1
      6: f1_0_main_Load -> [3] : [ arg1>0 && arg2>4 ], cost: NONTERM
      7: f1_0_main_Load -> f53_0_loop_EQ : arg1'=0, arg2'=arg2P_3, [ arg1>0 && arg2<5 && arg2>=1 ], cost: 1+arg2
      3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
      6: f1_0_main_Load -> [3] : [ arg1>0 && arg2>4 ], cost: NONTERM
      7: f1_0_main_Load -> f53_0_loop_EQ : arg1'=0, arg2'=arg2P_3, [ arg1>0 && arg2<5 && arg2>=1 ], cost: 1+arg2
      3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1

Eliminated locations (on tree-shaped paths):
   Start location: __init
      8: __init -> [3] : [ arg1P_4>0 && arg2P_4>4 ], cost: NONTERM
      9: __init -> f53_0_loop_EQ : arg1'=0, arg2'=arg2P_3, [ arg1P_4>0 && arg2P_4<5 && arg2P_4>=1 ], cost: 2+arg2P_4

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
      8: __init -> [3] : [ arg1P_4>0 && arg2P_4>4 ], cost: NONTERM
      9: __init -> f53_0_loop_EQ : arg1'=0, arg2'=arg2P_3, [ arg1P_4>0 && arg2P_4<5 && arg2P_4>=1 ], cost: 2+arg2P_4

Computing asymptotic complexity for rule 8
   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:  [ arg1P_4>0 && arg2P_4>4 ]

NO