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
      8: f3 -> f4 : arg1'=arg1P_9, arg2'=arg2P_9, [ arg1>=0 && arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1
      9: f3 -> f4 : arg1'=arg1P_10, arg2'=arg2P_10, [ arg2>=0 && arg1==arg1P_10 && arg2==arg2P_10 ], cost: 1
     11: f3 -> f10 : arg1'=arg1P_12, arg2'=arg2P_12, [ arg1<0 && arg2<0 && arg1==arg1P_12 && arg2==arg2P_12 ], cost: 1
      2: f5 -> f8 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1P_3==-1+arg1 && arg2==arg2P_3 ], cost: 1
      6: f8 -> f7 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1
      3: f6 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2P_4==-1+arg2 && arg1==arg1P_4 ], cost: 1
      7: f9 -> f7 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1==arg1P_8 && arg2==arg2P_8 ], cost: 1
      4: f4 -> f5 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1>=0 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      5: f4 -> f6 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1<0 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
     10: f7 -> f3 : arg1'=arg1P_11, arg2'=arg2P_11, [ arg1==arg1P_11 && arg2==arg2P_11 ], cost: 1
     12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
     12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, [], 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
      8: f3 -> f4 : arg1'=arg1P_9, arg2'=arg2P_9, [ arg1>=0 && arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1
      9: f3 -> f4 : arg1'=arg1P_10, arg2'=arg2P_10, [ arg2>=0 && arg1==arg1P_10 && arg2==arg2P_10 ], cost: 1
      2: f5 -> f8 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1P_3==-1+arg1 && arg2==arg2P_3 ], cost: 1
      6: f8 -> f7 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1
      3: f6 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2P_4==-1+arg2 && arg1==arg1P_4 ], cost: 1
      7: f9 -> f7 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1==arg1P_8 && arg2==arg2P_8 ], cost: 1
      4: f4 -> f5 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1>=0 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      5: f4 -> f6 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1<0 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
     10: f7 -> f3 : arg1'=arg1P_11, arg2'=arg2P_11, [ arg1==arg1P_11 && arg2==arg2P_11 ], cost: 1
     12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, [], 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
      8: f3 -> f4 : [ arg1>=0 ], cost: 1
      9: f3 -> f4 : [ arg2>=0 ], cost: 1
      2: f5 -> f8 : arg1'=-1+arg1, [], cost: 1
      6: f8 -> f7 : [], cost: 1
      3: f6 -> f9 : arg2'=-1+arg2, [], cost: 1
      7: f9 -> f7 : [], cost: 1
      4: f4 -> f5 : [ arg1>=0 ], cost: 1
      5: f4 -> f6 : [ arg1<0 ], cost: 1
     10: f7 -> f3 : [], cost: 1
     12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
      8: f3 -> f4 : [ arg1>=0 ], cost: 1
      9: f3 -> f4 : [ arg2>=0 ], cost: 1
     17: f4 -> f7 : arg1'=-1+arg1, [ arg1>=0 ], cost: 3
     18: f4 -> f7 : arg2'=-1+arg2, [ arg1<0 ], cost: 3
     10: f7 -> f3 : [], cost: 1
     14: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3

Eliminated locations (on tree-shaped paths):
   Start location: __init
     19: f3 -> f7 : arg1'=-1+arg1, [ arg1>=0 ], cost: 4
     20: f3 -> f7 : arg1'=-1+arg1, [ arg2>=0 && arg1>=0 ], cost: 4
     21: f3 -> f7 : arg2'=-1+arg2, [ arg2>=0 && arg1<0 ], cost: 4
     10: f7 -> f3 : [], cost: 1
     14: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3

Eliminated locations (on tree-shaped paths):
   Start location: __init
     22: f3 -> f3 : arg1'=-1+arg1, [ arg1>=0 ], cost: 5
     23: f3 -> f3 : arg1'=-1+arg1, [ arg2>=0 && arg1>=0 ], cost: 5
     24: f3 -> f3 : arg2'=-1+arg2, [ arg2>=0 && arg1<0 ], cost: 5
     14: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3

Accelerating simple loops of location 2.
   Accelerating the following rules:
     22: f3 -> f3 : arg1'=-1+arg1, [ arg1>=0 ], cost: 5
     23: f3 -> f3 : arg1'=-1+arg1, [ arg2>=0 && arg1>=0 ], cost: 5
     24: f3 -> f3 : arg2'=-1+arg2, [ arg2>=0 && arg1<0 ], cost: 5

   Accelerated rule 22 with backward acceleration, yielding the new rule 25.
   Accelerated rule 23 with backward acceleration, yielding the new rule 26.
   Accelerated rule 24 with backward acceleration, yielding the new rule 27.
[accelerate] Nesting with 3 inner and 3 outer candidates
   Removing the simple loops: 22 23 24.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     25: f3 -> f3 : arg1'=-1, [ 1+arg1>=0 ], cost: 5+5*arg1
     26: f3 -> f3 : arg1'=-1, [ arg2>=0 && 1+arg1>=0 ], cost: 5+5*arg1
     27: f3 -> f3 : arg2'=-1, [ arg1<0 && 1+arg2>=0 ], cost: 5+5*arg2
     14: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3

Chained accelerated rules (with incoming rules):
   Start location: __init
     14: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3
     28: __init -> f3 : arg1'=-1, arg2'=arg2P_2, [ 1+arg1P_1>=0 ], cost: 8+5*arg1P_1
     29: __init -> f3 : arg1'=-1, arg2'=arg2P_2, [ arg2P_2>=0 && 1+arg1P_1>=0 ], cost: 8+5*arg1P_1
     30: __init -> f3 : arg1'=arg1P_1, arg2'=-1, [ arg1P_1<0 && 1+arg2P_2>=0 ], cost: 8+5*arg2P_2

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     28: __init -> f3 : arg1'=-1, arg2'=arg2P_2, [ 1+arg1P_1>=0 ], cost: 8+5*arg1P_1
     29: __init -> f3 : arg1'=-1, arg2'=arg2P_2, [ arg2P_2>=0 && 1+arg1P_1>=0 ], cost: 8+5*arg1P_1
     30: __init -> f3 : arg1'=arg1P_1, arg2'=-1, [ arg1P_1<0 && 1+arg2P_2>=0 ], cost: 8+5*arg2P_2

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     28: __init -> f3 : arg1'=-1, arg2'=arg2P_2, [ 1+arg1P_1>=0 ], cost: 8+5*arg1P_1
     29: __init -> f3 : arg1'=-1, arg2'=arg2P_2, [ arg2P_2>=0 && 1+arg1P_1>=0 ], cost: 8+5*arg1P_1
     30: __init -> f3 : arg1'=arg1P_1, arg2'=-1, [ arg1P_1<0 && 1+arg2P_2>=0 ], cost: 8+5*arg2P_2

Computing asymptotic complexity for rule 28
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 29
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 30
   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),?)