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 && 1==arg2P_2 && arg3==arg3P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && 1==arg3P_3 ], cost: 1
      5: f4 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg3<arg1 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      6: f4 -> f5 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg3>arg1 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      8: f4 -> f8 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg3==arg1 && arg1==arg1P_9 && arg2==arg2P_9 && arg3==arg3P_9 ], 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
      4: f6 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2P_5==1+arg2 && arg1==arg1P_5 && arg3==arg3P_5 ], cost: 1
      7: f7 -> f4 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1
      9: __init -> f1 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      9: __init -> f1 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [], 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 && 1==arg2P_2 && arg3==arg3P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && 1==arg3P_3 ], cost: 1
      5: f4 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg3<arg1 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      6: f4 -> f5 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg3>arg1 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], 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
      4: f6 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2P_5==1+arg2 && arg1==arg1P_5 && arg3==arg3P_5 ], cost: 1
      7: f7 -> f4 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1
      9: __init -> f1 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1
      1: f2 -> f3 : arg2'=1, [], cost: 1
      2: f3 -> f4 : arg3'=1, [], cost: 1
      5: f4 -> f5 : [ arg3<arg1 ], cost: 1
      6: f4 -> f5 : [ arg3>arg1 ], cost: 1
      3: f5 -> f6 : arg3'=arg2*arg3, [], cost: 1
      4: f6 -> f7 : arg2'=1+arg2, [], cost: 1
      7: f7 -> f4 : [], cost: 1
      9: __init -> f1 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
      5: f4 -> f5 : [ arg3<arg1 ], cost: 1
      6: f4 -> f5 : [ arg3>arg1 ], cost: 1
     14: f5 -> f4 : arg2'=1+arg2, arg3'=arg2*arg3, [], cost: 3
     12: __init -> f4 : arg1'=arg1P_1, arg2'=1, arg3'=1, [], cost: 4

Eliminated locations (on tree-shaped paths):
   Start location: __init
     15: f4 -> f4 : arg2'=1+arg2, arg3'=arg2*arg3, [ arg3<arg1 ], cost: 4
     16: f4 -> f4 : arg2'=1+arg2, arg3'=arg2*arg3, [ arg3>arg1 ], cost: 4
     12: __init -> f4 : arg1'=arg1P_1, arg2'=1, arg3'=1, [], cost: 4

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

   Failed to prove monotonicity of the guard of rule 15.
   Failed to prove monotonicity of the guard of rule 16.
[accelerate] Nesting with 2 inner and 2 outer candidates

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     15: f4 -> f4 : arg2'=1+arg2, arg3'=arg2*arg3, [ arg3<arg1 ], cost: 4
     16: f4 -> f4 : arg2'=1+arg2, arg3'=arg2*arg3, [ arg3>arg1 ], cost: 4
     12: __init -> f4 : arg1'=arg1P_1, arg2'=1, arg3'=1, [], cost: 4

Chained accelerated rules (with incoming rules):
   Start location: __init
     12: __init -> f4 : arg1'=arg1P_1, arg2'=1, arg3'=1, [], cost: 4
     17: __init -> f4 : arg1'=arg1P_1, arg2'=2, arg3'=1, [ 1<arg1P_1 ], cost: 8
     18: __init -> f4 : arg1'=arg1P_1, arg2'=2, arg3'=1, [ 1>arg1P_1 ], cost: 8

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),?)