WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: __init
      0: f1_0_main_Load -> f42_0_decrease_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1>0 && arg2>-1 && arg2==arg1P_1 ], cost: 1
      1: f42_0_decrease_LE -> f42_0_decrease_LE : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1>5 && -1+arg1==arg1P_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 -> f42_0_decrease_LE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1
      1: f42_0_decrease_LE -> f42_0_decrease_LE : arg1'=-1+arg1, arg2'=arg2P_2, [ arg1>5 ], 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: f42_0_decrease_LE -> f42_0_decrease_LE : arg1'=-1+arg1, arg2'=arg2P_2, [ arg1>5 ], 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 -> f42_0_decrease_LE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1
      3: f42_0_decrease_LE -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ -5+arg1>=1 ], cost: -5+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 -> f42_0_decrease_LE : arg1'=arg2, arg2'=arg2P_1, [ arg1>0 && arg2>-1 ], cost: 1
      4: f1_0_main_Load -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ arg1>0 && -5+arg2>=1 ], cost: -4+arg2
      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 -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ arg1>0 && -5+arg2>=1 ], cost: -4+arg2
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1

Eliminated locations (on linear paths):
   Start location: __init
      5: __init -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ arg1P_3>0 && -5+arg2P_3>=1 ], cost: -3+arg2P_3

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
      5: __init -> f42_0_decrease_LE : arg1'=5, arg2'=arg2P_2, [ arg1P_3>0 && -5+arg2P_3>=1 ], cost: -3+arg2P_3

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),?)