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
      3: f2 -> f3 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg1>0 && 2*arg1<=arg2 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      5: f2 -> f6 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1<=0 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      6: f2 -> f6 : arg1'=arg1P_7, arg2'=arg2P_7, [ 2*arg1>arg2 && arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1
      1: f3 -> f4 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 && arg1==arg2P_2 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2==arg2P_3 ], cost: 1
      4: f5 -> f2 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], 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
      3: f2 -> f3 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg1>0 && 2*arg1<=arg2 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      1: f3 -> f4 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 && arg1==arg2P_2 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2==arg2P_3 ], cost: 1
      4: f5 -> f2 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], 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
      3: f2 -> f3 : [ arg1>0 && 2*arg1<=arg2 ], cost: 1
      1: f3 -> f4 : arg2'=arg1, [], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, [], cost: 1
      4: f5 -> f2 : [], 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
     11: f2 -> f2 : arg1'=arg1P_3, arg2'=arg1, [ arg1>0 && 2*arg1<=arg2 ], cost: 4
      8: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_8, [], cost: 2

Accelerating simple loops of location 1.
   Accelerating the following rules:
     11: f2 -> f2 : arg1'=arg1P_3, arg2'=arg1, [ arg1>0 && 2*arg1<=arg2 ], cost: 4

[test] deduced pseudo-invariant -1+2*arg1P_3<=0, also trying 1-2*arg1P_3<=-1
[test] deduced pseudo-invariant -arg2-4*arg1P_3+4*arg1<=0, also trying arg2+4*arg1P_3-4*arg1<=-1
   Accelerated rule 11 with backward acceleration, yielding the new rule 12.
[accelerate] Nesting with 1 inner and 1 outer candidates

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     11: f2 -> f2 : arg1'=arg1P_3, arg2'=arg1, [ arg1>0 && 2*arg1<=arg2 ], cost: 4
     12: f2 -> f2 : arg1'=arg1P_3, arg2'=arg1P_3, [ arg1>0 && 1-2*arg1P_3<=-1 && -arg2-4*arg1P_3+4*arg1<=0 && k>=1 && 2*arg1P_3<=arg1P_3 ], cost: 4*k
      8: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_8, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: __init
      8: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_8, [], cost: 2
     13: __init -> f2 : arg1'=arg1P_3, arg2'=arg1P_1, [ arg1P_1>0 ], cost: 6

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     <empty>

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     <empty>

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),?)