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, [ arg1==arg1P_1 && arg3==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg2==arg2P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ 0==arg1P_3 && arg2==arg2P_3 && arg3==arg3P_3 ], cost: 1
      5: f4 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg2>arg3 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      7: f4 -> f8 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg2<=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==1+arg3 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      4: f6 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg1P_5==1+arg1 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      6: f7 -> f4 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], 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, [ arg1==arg1P_1 && arg3==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg2==arg2P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ 0==arg1P_3 && arg2==arg2P_3 && arg3==arg3P_3 ], cost: 1
      5: f4 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg2>arg3 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg3P_4==1+arg3 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      4: f6 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg1P_5==1+arg1 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      6: f7 -> f4 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], 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 : arg2'=arg2P_1, [], cost: 1
      1: f2 -> f3 : arg3'=arg3P_2, [], cost: 1
      2: f3 -> f4 : arg1'=0, [], cost: 1
      5: f4 -> f5 : [ arg2>arg3 ], cost: 1
      3: f5 -> f6 : arg3'=1+arg3, [], cost: 1
      4: f6 -> f7 : arg1'=1+arg1, [], cost: 1
      6: f7 -> 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
     14: f4 -> f4 : arg1'=1+arg1, arg3'=1+arg3, [ arg2>arg3 ], cost: 4
     11: __init -> f4 : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_2, [], cost: 4

Accelerating simple loops of location 3.
   Accelerating the following rules:
     14: f4 -> f4 : arg1'=1+arg1, arg3'=1+arg3, [ arg2>arg3 ], cost: 4

   Accelerated rule 14 with backward acceleration, yielding the new rule 15.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 14.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     15: f4 -> f4 : arg1'=arg2-arg3+arg1, arg3'=arg2, [ arg2-arg3>=0 ], cost: 4*arg2-4*arg3
     11: __init -> f4 : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_2, [], cost: 4

Chained accelerated rules (with incoming rules):
   Start location: __init
     11: __init -> f4 : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_2, [], cost: 4
     16: __init -> f4 : arg1'=arg2P_1-arg3P_2, arg2'=arg2P_1, arg3'=arg2P_1, [ arg2P_1-arg3P_2>=0 ], cost: 4+4*arg2P_1-4*arg3P_2

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     16: __init -> f4 : arg1'=arg2P_1-arg3P_2, arg2'=arg2P_1, arg3'=arg2P_1, [ arg2P_1-arg3P_2>=0 ], cost: 4+4*arg2P_1-4*arg3P_2

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     16: __init -> f4 : arg1'=arg2P_1-arg3P_2, arg2'=arg2P_1, arg3'=arg2P_1, [ arg2P_1-arg3P_2>=0 ], cost: 4+4*arg2P_1-4*arg3P_2

Computing asymptotic complexity for rule 16
   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),?)