WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: __init
      0: f1_0_main_Load -> f145_0_main_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1
      1: f145_0_main_LE -> f145_0_main_LE : arg1'=arg1P_2, arg2'=arg2P_2, [ arg2<arg1 && -1+arg1==arg1P_2 && arg2==arg2P_2 ], cost: 1
      2: __init -> f1_0_main_Load : 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_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1_0_main_Load -> f145_0_main_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1
      1: f145_0_main_LE -> f145_0_main_LE : arg1'=-1+arg1, [ arg2<arg1 ], cost: 1
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

### Simplification by acceleration and chaining ###

Accelerating simple loops of location 1.
   Accelerating the following rules:
      1: f145_0_main_LE -> f145_0_main_LE : arg1'=-1+arg1, [ arg2<arg1 ], cost: 1

   Accelerated rule 1 with backward acceleration, yielding the new rule 3.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 1.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
      0: f1_0_main_Load -> f145_0_main_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1
      3: f145_0_main_LE -> f145_0_main_LE : arg1'=arg2, [ -arg2+arg1>=0 ], cost: -arg2+arg1
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Chained accelerated rules (with incoming rules):
   Start location: __init
      0: f1_0_main_Load -> f145_0_main_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1
      4: f1_0_main_Load -> f145_0_main_LE : arg1'=arg2P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 && arg1P_1-arg2P_1>=0 ], cost: 1+arg1P_1-arg2P_1
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
      4: f1_0_main_Load -> f145_0_main_LE : arg1'=arg2P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 && arg1P_1-arg2P_1>=0 ], cost: 1+arg1P_1-arg2P_1
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Eliminated locations (on linear paths):
   Start location: __init
      5: __init -> f145_0_main_LE : arg1'=arg2P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2P_3>-1 && arg1P_1>-1 && arg1P_3>0 && arg1P_1-arg2P_1>=0 ], cost: 2+arg1P_1-arg2P_1

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
      5: __init -> f145_0_main_LE : arg1'=arg2P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2P_3>-1 && arg1P_1>-1 && arg1P_3>0 && arg1P_1-arg2P_1>=0 ], cost: 2+arg1P_1-arg2P_1

Computing asymptotic complexity for rule 5
   Resulting cost 0 has complexity: Unknown

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),?)