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, [ arg1==arg1P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg2==arg2P_2 ], cost: 1
      7: f3 -> f4 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1>=2 && arg1==arg1P_8 && arg2==arg2P_8 ], cost: 1
      8: f3 -> f5 : arg1'=arg1P_9, arg2'=arg2P_9, [ arg1<2 && arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1
      2: f7 -> f8 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==arg2-arg1 && arg1==arg1P_3 ], cost: 1
      3: f8 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg1P_4==1+arg1 && arg2==arg2P_4 ], cost: 1
      5: f9 -> f4 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      4: f4 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg2+arg1>=0 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      6: f4 -> f10 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg2+arg1<0 && arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1
      9: f10 -> f6 : arg1'=arg1P_10, arg2'=arg2P_10, [ arg1==arg1P_10 && arg2==arg2P_10 ], cost: 1
     10: f5 -> f6 : arg1'=arg1P_11, arg2'=arg2P_11, [ arg1==arg1P_11 && arg2==arg2P_11 ], cost: 1
     11: __init -> f1 : arg1'=arg1P_12, arg2'=arg2P_12, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
     11: __init -> f1 : arg1'=arg1P_12, arg2'=arg2P_12, [], cost: 1

Removed unreachable and leaf rules:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1==arg1P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg2==arg2P_2 ], cost: 1
      7: f3 -> f4 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1>=2 && arg1==arg1P_8 && arg2==arg2P_8 ], cost: 1
      2: f7 -> f8 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==arg2-arg1 && arg1==arg1P_3 ], cost: 1
      3: f8 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg1P_4==1+arg1 && arg2==arg2P_4 ], cost: 1
      5: f9 -> f4 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      4: f4 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg2+arg1>=0 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
     11: __init -> f1 : arg1'=arg1P_12, arg2'=arg2P_12, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg2'=arg2P_1, [], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, [], cost: 1
      7: f3 -> f4 : [ arg1>=2 ], cost: 1
      2: f7 -> f8 : arg2'=arg2-arg1, [], cost: 1
      3: f8 -> f9 : arg1'=1+arg1, [], cost: 1
      5: f9 -> f4 : [], cost: 1
      4: f4 -> f7 : [ arg2+arg1>=0 ], cost: 1
     11: __init -> f1 : arg1'=arg1P_12, arg2'=arg2P_12, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     17: f4 -> f4 : arg1'=1+arg1, arg2'=arg2-arg1, [ arg2+arg1>=0 ], cost: 4
     14: __init -> f4 : arg1'=arg1P_2, arg2'=arg2P_1, [ arg1P_2>=2 ], cost: 4

Accelerating simple loops of location 6.
   Accelerating the following rules:
     17: f4 -> f4 : arg1'=1+arg1, arg2'=arg2-arg1, [ arg2+arg1>=0 ], cost: 4

[test] deduced invariant 1-arg1<=0
   Accelerated rule 17 with backward acceleration, yielding the new rule 18.
[accelerate] Nesting with 1 inner and 1 outer candidates

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     17: f4 -> f4 : arg1'=1+arg1, arg2'=arg2-arg1, [ arg2+arg1>=0 ], cost: 4
     18: f4 -> f4 : arg1'=k+arg1, arg2'=arg2+1/2*k-k*arg1-1/2*k^2, [ 1-arg1<=0 && k>=0 && -3/2+arg2-1/2*(-1+k)^2+3/2*k-(-1+k)*arg1+arg1>=0 ], cost: 4*k
     14: __init -> f4 : arg1'=arg1P_2, arg2'=arg2P_1, [ arg1P_2>=2 ], cost: 4

Chained accelerated rules (with incoming rules):
   Start location: __init
     14: __init -> f4 : arg1'=arg1P_2, arg2'=arg2P_1, [ arg1P_2>=2 ], cost: 4
     19: __init -> f4 : arg1'=1+arg1P_2, arg2'=arg2P_1-arg1P_2, [ arg1P_2>=2 && arg2P_1+arg1P_2>=0 ], cost: 8
     20: __init -> f4 : arg1'=k+arg1P_2, arg2'=1/2*k+arg2P_1-1/2*k^2-k*arg1P_2, [ arg1P_2>=2 && k>=0 && -3/2-1/2*(-1+k)^2+3/2*k+arg2P_1+arg1P_2-(-1+k)*arg1P_2>=0 ], cost: 4+4*k

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     20: __init -> f4 : arg1'=k+arg1P_2, arg2'=1/2*k+arg2P_1-1/2*k^2-k*arg1P_2, [ arg1P_2>=2 && k>=0 && -3/2-1/2*(-1+k)^2+3/2*k+arg2P_1+arg1P_2-(-1+k)*arg1P_2>=0 ], cost: 4+4*k

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     20: __init -> f4 : arg1'=k+arg1P_2, arg2'=1/2*k+arg2P_1-1/2*k^2-k*arg1P_2, [ arg1P_2>=2 && k>=0 && -3/2-1/2*(-1+k)^2+3/2*k+arg2P_1+arg1P_2-(-1+k)*arg1P_2>=0 ], cost: 4+4*k

Computing asymptotic complexity for rule 20
   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),?)