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
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 ], cost: 1
      4: f3 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1>arg2 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      6: f3 -> f7 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1<=arg2 && arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1P_3==1+arg1 && arg2==arg2P_3 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2P_4==2+arg2 && arg1==arg1P_4 ], cost: 1
      5: f6 -> f3 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

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

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     12: f3 -> f3 : arg1'=1+arg1, arg2'=2+arg2, [ arg1>arg2 ], cost: 4
      9: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3

Accelerating simple loops of location 2.
   Accelerating the following rules:
     12: f3 -> f3 : arg1'=1+arg1, arg2'=2+arg2, [ arg1>arg2 ], cost: 4

   Accelerated rule 12 with backward acceleration, yielding the new rule 13.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 12.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     13: f3 -> f3 : arg1'=-arg2+2*arg1, arg2'=-arg2+2*arg1, [ -arg2+arg1>=0 ], cost: -4*arg2+4*arg1
      9: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3

Chained accelerated rules (with incoming rules):
   Start location: __init
      9: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3
     14: __init -> f3 : arg1'=-arg2P_2+2*arg1P_1, arg2'=-arg2P_2+2*arg1P_1, [ -arg2P_2+arg1P_1>=0 ], cost: 3-4*arg2P_2+4*arg1P_1

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     14: __init -> f3 : arg1'=-arg2P_2+2*arg1P_1, arg2'=-arg2P_2+2*arg1P_1, [ -arg2P_2+arg1P_1>=0 ], cost: 3-4*arg2P_2+4*arg1P_1

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     14: __init -> f3 : arg1'=-arg2P_2+2*arg1P_1, arg2'=-arg2P_2+2*arg1P_1, [ -arg2P_2+arg1P_1>=0 ], cost: 3-4*arg2P_2+4*arg1P_1

Computing asymptotic complexity for rule 14
   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),?)