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 && 2==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
      8: f4 -> f5 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg1<0 && arg1==arg1P_9 && arg2==arg2P_9 && arg3==arg3P_9 ], cost: 1
      9: f4 -> f5 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg2<1 && arg1==arg1P_10 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1
     10: f4 -> f6 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg1>=0 && arg2>=1 && arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1
      3: f8 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg2P_4==arg2^2 && arg1==arg1P_4 && arg3==arg3P_4 ], cost: 1
      4: f9 -> f10 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg3P_5==2*arg3 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      6: f10 -> f6 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      5: f6 -> f8 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1>arg2 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      7: f6 -> f11 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1<=arg2 && arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1
     12: f11 -> f7 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [ arg1==arg1P_13 && arg2==arg2P_13 && arg3==arg3P_13 ], cost: 1
     11: f5 -> f7 : arg1'=arg1P_12, arg2'=arg2P_12, arg3'=arg3P_12, [ arg1==arg1P_12 && arg2==arg2P_12 && arg3==arg3P_12 ], cost: 1
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], 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 && 2==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
     10: f4 -> f6 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg1>=0 && arg2>=1 && arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1
      3: f8 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg2P_4==arg2^2 && arg1==arg1P_4 && arg3==arg3P_4 ], cost: 1
      4: f9 -> f10 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg3P_5==2*arg3 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      6: f10 -> f6 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      5: f6 -> f8 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1>arg2 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1
      1: f2 -> f3 : arg2'=2, [], cost: 1
      2: f3 -> f4 : arg3'=1, [], cost: 1
     10: f4 -> f6 : [ arg1>=0 && arg2>=1 ], cost: 1
      3: f8 -> f9 : arg2'=arg2^2, [], cost: 1
      4: f9 -> f10 : arg3'=2*arg3, [], cost: 1
      6: f10 -> f6 : [], cost: 1
      5: f6 -> f8 : [ arg1>arg2 ], cost: 1
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     20: f6 -> f6 : arg2'=arg2^2, arg3'=2*arg3, [ arg1>arg2 ], cost: 4
     17: __init -> f6 : arg1'=arg1P_1, arg2'=2, arg3'=1, [ arg1P_1>=0 ], cost: 5

Accelerating simple loops of location 7.
   Accelerating the following rules:
     20: f6 -> f6 : arg2'=arg2^2, arg3'=2*arg3, [ arg1>arg2 ], cost: 4

   Accelerated rule 20 with non-termination, yielding the new rule 21.
   Accelerated rule 20 with backward acceleration, yielding the new rule 22.
[accelerate] Nesting with 1 inner and 0 outer candidates
   Removing the simple loops: 20.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     21: f6 -> [12] : [ arg1>arg2 && arg2==0 && arg1==1 ], cost: NONTERM
     22: f6 -> f6 : arg2'=arg2^(2^k), arg3'=2^k*arg3, [ k>=0 && arg1>arg2^(2^(-1+k)) ], cost: 4*k
     17: __init -> f6 : arg1'=arg1P_1, arg2'=2, arg3'=1, [ arg1P_1>=0 ], cost: 5

Chained accelerated rules (with incoming rules):
   Start location: __init
     17: __init -> f6 : arg1'=arg1P_1, arg2'=2, arg3'=1, [ arg1P_1>=0 ], cost: 5
     23: __init -> f6 : arg1'=arg1P_1, arg2'=2^(2^k), arg3'=2^k, [ arg1P_1>=0 && k>=0 && arg1P_1>2^(2^(-1+k)) ], cost: 5+4*k

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     23: __init -> f6 : arg1'=arg1P_1, arg2'=2^(2^k), arg3'=2^k, [ arg1P_1>=0 && k>=0 && arg1P_1>2^(2^(-1+k)) ], cost: 5+4*k

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     23: __init -> f6 : arg1'=arg1P_1, arg2'=2^(2^k), arg3'=2^k, [ arg1P_1>=0 && k>=0 && arg1P_1>2^(2^(-1+k)) ], cost: 5+4*k

Computing asymptotic complexity for rule 23
   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),?)