WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: __init
      0: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg1>0 && 0==arg2 && 0==arg1P_1 && 0==arg2P_1 && 0==arg3P_1 ], cost: 1
      1: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1>0 && arg1P_2>-1 && 1==arg2 && 0==arg2P_2 && arg1P_2==arg3P_2 ], cost: 1
      2: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg2P_3>-1 && arg2>1 && arg1P_3>-1 && arg1>0 && arg1P_3==arg3P_3 ], cost: 1
      3: f264_0_loop_LE -> f264_0_loop_LE : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1>0 && -1+arg2<arg2 && arg1==arg3 && arg2+arg1==arg1P_4 && -1+arg2==arg2P_4 && arg2+arg1==arg3P_4 ], cost: 1
      4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1_0_main_Load -> f264_0_loop_LE : arg1'=0, arg2'=0, arg3'=0, [ arg1>0 && 0==arg2 ], cost: 1
      1: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg3P_2, arg2'=0, arg3'=arg3P_2, [ arg1>0 && arg3P_2>-1 && 1==arg2 ], cost: 1
      2: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg3P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg2P_3>-1 && arg2>1 && arg3P_3>-1 && arg1>0 ], cost: 1
      3: f264_0_loop_LE -> f264_0_loop_LE : arg1'=arg2+arg1, arg2'=-1+arg2, arg3'=arg2+arg1, [ arg1>0 && arg1==arg3 ], cost: 1
      4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [], cost: 1

### Simplification by acceleration and chaining ###

Accelerating simple loops of location 1.
   Accelerating the following rules:
      3: f264_0_loop_LE -> f264_0_loop_LE : arg1'=arg2+arg1, arg2'=-1+arg2, arg3'=arg2+arg1, [ arg1>0 && arg1==arg3 ], cost: 1

   Failed to prove monotonicity of the guard of rule 3.
[accelerate] Nesting with 1 inner and 1 outer candidates

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
      0: f1_0_main_Load -> f264_0_loop_LE : arg1'=0, arg2'=0, arg3'=0, [ arg1>0 && 0==arg2 ], cost: 1
      1: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg3P_2, arg2'=0, arg3'=arg3P_2, [ arg1>0 && arg3P_2>-1 && 1==arg2 ], cost: 1
      2: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg3P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg2P_3>-1 && arg2>1 && arg3P_3>-1 && arg1>0 ], cost: 1
      3: f264_0_loop_LE -> f264_0_loop_LE : arg1'=arg2+arg1, arg2'=-1+arg2, arg3'=arg2+arg1, [ arg1>0 && arg1==arg3 ], cost: 1
      4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [], cost: 1

Chained accelerated rules (with incoming rules):
   Start location: __init
      0: f1_0_main_Load -> f264_0_loop_LE : arg1'=0, arg2'=0, arg3'=0, [ arg1>0 && 0==arg2 ], cost: 1
      1: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg3P_2, arg2'=0, arg3'=arg3P_2, [ arg1>0 && arg3P_2>-1 && 1==arg2 ], cost: 1
      2: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg3P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg2P_3>-1 && arg2>1 && arg3P_3>-1 && arg1>0 ], cost: 1
      5: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg3P_2, arg2'=-1, arg3'=arg3P_2, [ arg1>0 && 1==arg2 && arg3P_2>0 ], cost: 2
      6: f1_0_main_Load -> f264_0_loop_LE : arg1'=arg2P_3+arg3P_3, arg2'=-1+arg2P_3, arg3'=arg2P_3+arg3P_3, [ arg2P_3>-1 && arg2>1 && arg1>0 && arg3P_3>0 ], cost: 2
      4: __init -> f1_0_main_Load : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [], cost: 1

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     <empty>

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     <empty>

Obtained the following overall complexity (w.r.t. the length of the input n):
   Complexity:  Constant
   Cpx degree:  0
   Solved cost: 1
   Rule cost:   1
   Rule guard:  []

WORST_CASE(Omega(1),?)