WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: __init
      0: f1_0_main_Load -> f259_0_div_LT : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg3P_1>-1 && arg2>-1 && arg2P_1>-1 && arg1>0 && 0==arg1P_1 ], cost: 1
      1: f259_0_div_LT -> f259_0_div_LT : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg2>-1 && arg3>0 && arg3<=arg2 && arg1>-1 && 1+arg1==arg1P_2 && arg2-arg3==arg2P_2 && arg3==arg3P_2 ], cost: 1
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_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, arg3'=arg3P_3, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1_0_main_Load -> f259_0_div_LT : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_1, [ arg3P_1>-1 && arg2>-1 && arg2P_1>-1 && arg1>0 ], cost: 1
      1: f259_0_div_LT -> f259_0_div_LT : arg1'=1+arg1, arg2'=arg2-arg3, [ arg3>0 && arg3<=arg2 && arg1>-1 ], cost: 1
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

### Simplification by acceleration and chaining ###

Accelerating simple loops of location 1.
   Accelerating the following rules:
      1: f259_0_div_LT -> f259_0_div_LT : arg1'=1+arg1, arg2'=arg2-arg3, [ arg3>0 && arg3<=arg2 && arg1>-1 ], 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 -> f259_0_div_LT : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_1, [ arg3P_1>-1 && arg2>-1 && arg2P_1>-1 && arg1>0 ], cost: 1
      3: f259_0_div_LT -> f259_0_div_LT : arg1'=k+arg1, arg2'=arg2-arg3*k, [ arg3>0 && arg1>-1 && k>=0 && arg3<=arg2-(-1+k)*arg3 ], cost: k
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

Chained accelerated rules (with incoming rules):
   Start location: __init
      0: f1_0_main_Load -> f259_0_div_LT : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_1, [ arg3P_1>-1 && arg2>-1 && arg2P_1>-1 && arg1>0 ], cost: 1
      4: f1_0_main_Load -> f259_0_div_LT : arg1'=k, arg2'=-arg3P_1*k+arg2P_1, arg3'=arg3P_1, [ arg2>-1 && arg2P_1>-1 && arg1>0 && arg3P_1>0 && k>=0 && arg3P_1<=-arg3P_1*(-1+k)+arg2P_1 ], cost: 1+k
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
      4: f1_0_main_Load -> f259_0_div_LT : arg1'=k, arg2'=-arg3P_1*k+arg2P_1, arg3'=arg3P_1, [ arg2>-1 && arg2P_1>-1 && arg1>0 && arg3P_1>0 && k>=0 && arg3P_1<=-arg3P_1*(-1+k)+arg2P_1 ], cost: 1+k
      2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

Eliminated locations (on linear paths):
   Start location: __init
      5: __init -> f259_0_div_LT : arg1'=k, arg2'=-arg3P_1*k+arg2P_1, arg3'=arg3P_1, [ arg2P_3>-1 && arg2P_1>-1 && arg1P_3>0 && arg3P_1>0 && k>=0 && arg3P_1<=-arg3P_1*(-1+k)+arg2P_1 ], cost: 2+k

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
      5: __init -> f259_0_div_LT : arg1'=k, arg2'=-arg3P_1*k+arg2P_1, arg3'=arg3P_1, [ arg2P_3>-1 && arg2P_1>-1 && arg1P_3>0 && arg3P_1>0 && k>=0 && arg3P_1<=-arg3P_1*(-1+k)+arg2P_1 ], cost: 2+k

Computing asymptotic complexity for rule 5
   Simplified the guard:
      5: __init -> f259_0_div_LT : arg1'=k, arg2'=-arg3P_1*k+arg2P_1, arg3'=arg3P_1, [ arg2P_3>-1 && arg1P_3>0 && arg3P_1>0 && k>=0 && arg3P_1<=-arg3P_1*(-1+k)+arg2P_1 ], cost: 2+k
   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),?)