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, [ arg2==arg2P_1 ], cost: 1
      7: f2 -> f3 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1>0 && arg1==arg1P_8 && arg2==arg2P_8 ], cost: 1
      9: f2 -> f9 : arg1'=arg1P_10, arg2'=arg2P_10, [ arg1<=0 && arg1==arg1P_10 && arg2==arg2P_10 ], cost: 1
      1: f3 -> f4 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 && 0==arg2P_2 ], cost: 1
      3: f4 -> f5 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2<arg1 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      5: f4 -> f7 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg2>=arg1 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      2: f5 -> f6 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==1+arg2 && arg1==arg1P_3 ], cost: 1
      4: f6 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      6: f7 -> f8 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1P_7==-1+arg1 && arg2==arg2P_7 ], cost: 1
      8: f8 -> f2 : arg1'=arg1P_9, arg2'=arg2P_9, [ arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1
     10: __init -> f1 : arg1'=arg1P_11, arg2'=arg2P_11, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
     10: __init -> f1 : arg1'=arg1P_11, arg2'=arg2P_11, [], cost: 1

Removed unreachable and leaf rules:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2==arg2P_1 ], cost: 1
      7: f2 -> f3 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1>0 && arg1==arg1P_8 && arg2==arg2P_8 ], cost: 1
      1: f3 -> f4 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 && 0==arg2P_2 ], cost: 1
      3: f4 -> f5 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2<arg1 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      5: f4 -> f7 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg2>=arg1 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      2: f5 -> f6 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==1+arg2 && arg1==arg1P_3 ], cost: 1
      4: f6 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      6: f7 -> f8 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1P_7==-1+arg1 && arg2==arg2P_7 ], cost: 1
      8: f8 -> f2 : arg1'=arg1P_9, arg2'=arg2P_9, [ arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1
     10: __init -> f1 : arg1'=arg1P_11, arg2'=arg2P_11, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1
      7: f2 -> f3 : [ arg1>0 ], cost: 1
      1: f3 -> f4 : arg2'=0, [], cost: 1
      3: f4 -> f5 : [ arg2<arg1 ], cost: 1
      5: f4 -> f7 : [ arg2>=arg1 ], cost: 1
      2: f5 -> f6 : arg2'=1+arg2, [], cost: 1
      4: f6 -> f4 : [], cost: 1
      6: f7 -> f8 : arg1'=-1+arg1, [], cost: 1
      8: f8 -> f2 : [], cost: 1
     10: __init -> f1 : arg1'=arg1P_11, arg2'=arg2P_11, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     12: f2 -> f4 : arg2'=0, [ arg1>0 ], cost: 2
     15: f4 -> f4 : arg2'=1+arg2, [ arg2<arg1 ], cost: 3
     16: f4 -> f2 : arg1'=-1+arg1, [ arg2>=arg1 ], cost: 3
     11: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_11, [], cost: 2

Accelerating simple loops of location 3.
   Accelerating the following rules:
     15: f4 -> f4 : arg2'=1+arg2, [ arg2<arg1 ], cost: 3

   Accelerated rule 15 with backward acceleration, yielding the new rule 17.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 15.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     12: f2 -> f4 : arg2'=0, [ arg1>0 ], cost: 2
     16: f4 -> f2 : arg1'=-1+arg1, [ arg2>=arg1 ], cost: 3
     17: f4 -> f4 : arg2'=arg1, [ -arg2+arg1>=0 ], cost: -3*arg2+3*arg1
     11: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_11, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: __init
     12: f2 -> f4 : arg2'=0, [ arg1>0 ], cost: 2
     18: f2 -> f4 : arg2'=arg1, [ arg1>0 ], cost: 2+3*arg1
     16: f4 -> f2 : arg1'=-1+arg1, [ arg2>=arg1 ], cost: 3
     11: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_11, [], cost: 2

Eliminated locations (on tree-shaped paths):
   Start location: __init
     19: f2 -> f2 : arg1'=-1+arg1, arg2'=arg1, [ arg1>0 ], cost: 5+3*arg1
     11: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_11, [], cost: 2

Accelerating simple loops of location 1.
   Accelerating the following rules:
     19: f2 -> f2 : arg1'=-1+arg1, arg2'=arg1, [ arg1>0 ], cost: 5+3*arg1

   Accelerated rule 19 with backward acceleration, yielding the new rule 20.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 19.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     20: f2 -> f2 : arg1'=0, arg2'=1, [ arg1>=1 ], cost: 3/2*arg1^2+13/2*arg1
     11: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_11, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: __init
     11: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_11, [], cost: 2
     21: __init -> f2 : arg1'=0, arg2'=1, [ arg1P_1>=1 ], cost: 2+13/2*arg1P_1+3/2*arg1P_1^2

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     21: __init -> f2 : arg1'=0, arg2'=1, [ arg1P_1>=1 ], cost: 2+13/2*arg1P_1+3/2*arg1P_1^2

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     21: __init -> f2 : arg1'=0, arg2'=1, [ arg1P_1>=1 ], cost: 2+13/2*arg1P_1+3/2*arg1P_1^2

Computing asymptotic complexity for rule 21
   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),?)