NO

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

Initial linear ITS problem
   Start location: __init
      0: f1_0_main_ConstantStackPush -> f33_0_loop_GE : arg1'=arg1P_1, arg2'=arg2P_1, [ 1>-arg2 && arg2>-1 && arg1>0 && -arg2==arg1P_1 ], cost: 1
      1: f33_0_loop_GE -> f33_0_loop_GE : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1<1 && arg1<2 && -1+arg1<arg1 && arg1<0 && -1+arg1==arg1P_2 ], cost: 1
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1_0_main_ConstantStackPush -> f33_0_loop_GE : arg1'=-arg2, arg2'=arg2P_1, [ 1>-arg2 && arg1>0 ], cost: 1
      1: f33_0_loop_GE -> f33_0_loop_GE : arg1'=-1+arg1, arg2'=arg2P_2, [ arg1<0 ], cost: 1
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

### Simplification by acceleration and chaining ###

Accelerating simple loops of location 1.
   Accelerating the following rules:
      1: f33_0_loop_GE -> f33_0_loop_GE : arg1'=-1+arg1, arg2'=arg2P_2, [ arg1<0 ], cost: 1

   Accelerated rule 1 with non-termination, yielding the new rule 3.
[accelerate] Nesting with 0 inner and 0 outer candidates
   Removing the simple loops: 1.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
      0: f1_0_main_ConstantStackPush -> f33_0_loop_GE : arg1'=-arg2, arg2'=arg2P_1, [ 1>-arg2 && arg1>0 ], cost: 1
      3: f33_0_loop_GE -> [3] : [ arg1<0 ], cost: NONTERM
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Chained accelerated rules (with incoming rules):
   Start location: __init
      0: f1_0_main_ConstantStackPush -> f33_0_loop_GE : arg1'=-arg2, arg2'=arg2P_1, [ 1>-arg2 && arg1>0 ], cost: 1
      4: f1_0_main_ConstantStackPush -> [3] : [ arg1>0 && -arg2<0 ], cost: NONTERM
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
      4: f1_0_main_ConstantStackPush -> [3] : [ arg1>0 && -arg2<0 ], cost: NONTERM
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Eliminated locations (on linear paths):
   Start location: __init
      5: __init -> [3] : [ arg1P_3>0 && -arg2P_3<0 ], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
      5: __init -> [3] : [ arg1P_3>0 && -arg2P_3<0 ], cost: NONTERM

Computing asymptotic complexity for rule 5
   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_3>0 && -arg2P_3<0 ]

NO