WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg3==arg3P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 ], cost: 1
      4: f4 -> f5 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2>0 && arg1>=arg3 && arg1==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      6: f4 -> f7 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg2<=0 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      7: f4 -> f7 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1<arg3 && arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg3P_4==arg2+arg3 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      5: f6 -> f4 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      8: __init -> f1 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      8: __init -> f1 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [], cost: 1

Removed unreachable and leaf rules:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg3==arg3P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 ], cost: 1
      4: f4 -> f5 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2>0 && arg1>=arg3 && arg1==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg3P_4==arg2+arg3 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      5: f6 -> f4 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      8: __init -> f1 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1
      1: f2 -> f3 : arg2'=arg2P_2, [], cost: 1
      2: f3 -> f4 : arg3'=arg3P_3, [], cost: 1
      4: f4 -> f5 : [ arg2>0 && arg1>=arg3 ], cost: 1
      3: f5 -> f6 : arg3'=arg2+arg3, [], cost: 1
      5: f6 -> f4 : [], cost: 1
      8: __init -> f1 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     13: f4 -> f4 : arg3'=arg2+arg3, [ arg2>0 && arg1>=arg3 ], cost: 3
     11: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [], cost: 4

Accelerating simple loops of location 3.
   Accelerating the following rules:
     13: f4 -> f4 : arg3'=arg2+arg3, [ arg2>0 && arg1>=arg3 ], cost: 3

   Accelerated rule 13 with backward acceleration, yielding the new rule 14.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 13.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     14: f4 -> f4 : arg3'=arg3+arg2*k, [ arg2>0 && k>=0 && arg1>=arg2*(-1+k)+arg3 ], cost: 3*k
     11: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [], cost: 4

Chained accelerated rules (with incoming rules):
   Start location: __init
     11: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [], cost: 4
     15: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3+k*arg2P_2, [ arg2P_2>0 && k>=0 && arg1P_1>=arg2P_2*(-1+k)+arg3P_3 ], cost: 4+3*k

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     15: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3+k*arg2P_2, [ arg2P_2>0 && k>=0 && arg1P_1>=arg2P_2*(-1+k)+arg3P_3 ], cost: 4+3*k

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     15: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3+k*arg2P_2, [ arg2P_2>0 && k>=0 && arg1P_1>=arg2P_2*(-1+k)+arg3P_3 ], cost: 4+3*k

Computing asymptotic complexity for rule 15
   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),?)