WORST_CASE(Omega(n^1),?)

### Pre-processing the ITS problem ###

Initial linear ITS problem
   Start location: l4
      0: l0 -> l1 : __const_20^0'=__const_20^post_1, i^0'=i^post_1, j^0'=j^post_1, [ __const_20^0<=j^0 && __const_20^0==__const_20^post_1 && i^0==i^post_1 && j^0==j^post_1 ], cost: 1
      1: l0 -> l2 : __const_20^0'=__const_20^post_2, i^0'=i^post_2, j^0'=j^post_2, [ 1+j^0<=__const_20^0 && __const_20^0==__const_20^post_2 && i^0==i^post_2 && j^0==j^post_2 ], cost: 1
      2: l2 -> l0 : __const_20^0'=__const_20^post_3, i^0'=i^post_3, j^0'=j^post_3, [ j^post_3==2+j^0 && i^post_3==2+j^post_3 && __const_20^0==__const_20^post_3 ], cost: 1
      3: l3 -> l2 : __const_20^0'=__const_20^post_4, i^0'=i^post_4, j^0'=j^post_4, [ j^post_4==0 && __const_20^0==__const_20^post_4 && i^0==i^post_4 ], cost: 1
      4: l4 -> l3 : __const_20^0'=__const_20^post_5, i^0'=i^post_5, j^0'=j^post_5, [ __const_20^0==__const_20^post_5 && i^0==i^post_5 && j^0==j^post_5 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      4: l4 -> l3 : __const_20^0'=__const_20^post_5, i^0'=i^post_5, j^0'=j^post_5, [ __const_20^0==__const_20^post_5 && i^0==i^post_5 && j^0==j^post_5 ], cost: 1

Removed unreachable and leaf rules:
   Start location: l4
      1: l0 -> l2 : __const_20^0'=__const_20^post_2, i^0'=i^post_2, j^0'=j^post_2, [ 1+j^0<=__const_20^0 && __const_20^0==__const_20^post_2 && i^0==i^post_2 && j^0==j^post_2 ], cost: 1
      2: l2 -> l0 : __const_20^0'=__const_20^post_3, i^0'=i^post_3, j^0'=j^post_3, [ j^post_3==2+j^0 && i^post_3==2+j^post_3 && __const_20^0==__const_20^post_3 ], cost: 1
      3: l3 -> l2 : __const_20^0'=__const_20^post_4, i^0'=i^post_4, j^0'=j^post_4, [ j^post_4==0 && __const_20^0==__const_20^post_4 && i^0==i^post_4 ], cost: 1
      4: l4 -> l3 : __const_20^0'=__const_20^post_5, i^0'=i^post_5, j^0'=j^post_5, [ __const_20^0==__const_20^post_5 && i^0==i^post_5 && j^0==j^post_5 ], cost: 1

Simplified all rules, resulting in:
   Start location: l4
      1: l0 -> l2 : [ 1+j^0<=__const_20^0 ], cost: 1
      2: l2 -> l0 : i^0'=4+j^0, j^0'=2+j^0, [], cost: 1
      3: l3 -> l2 : j^0'=0, [], cost: 1
      4: l4 -> l3 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l4
      6: l2 -> l2 : i^0'=4+j^0, j^0'=2+j^0, [ 3+j^0<=__const_20^0 ], cost: 2
      5: l4 -> l2 : j^0'=0, [], cost: 2

Accelerating simple loops of location 2.
   Accelerating the following rules:
      6: l2 -> l2 : i^0'=4+j^0, j^0'=2+j^0, [ 3+j^0<=__const_20^0 ], cost: 2

   Accelerated rule 6 with metering function meter (where 2*meter==-2-j^0+__const_20^0), yielding the new rule 7.
   Removing the simple loops: 6.

Accelerated all simple loops using metering functions (where possible):
   Start location: l4
      7: l2 -> l2 : i^0'=2+j^0+2*meter, j^0'=j^0+2*meter, [ 3+j^0<=__const_20^0 && 2*meter==-2-j^0+__const_20^0 && meter>=1 ], cost: 2*meter
      5: l4 -> l2 : j^0'=0, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l4
      5: l4 -> l2 : j^0'=0, [], cost: 2
      8: l4 -> l2 : i^0'=2+2*meter, j^0'=2*meter, [ 3<=__const_20^0 && 2*meter==-2+__const_20^0 && meter>=1 ], cost: 2+2*meter

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l4
      8: l4 -> l2 : i^0'=2+2*meter, j^0'=2*meter, [ 3<=__const_20^0 && 2*meter==-2+__const_20^0 && meter>=1 ], cost: 2+2*meter

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l4
      8: l4 -> l2 : i^0'=2+2*meter, j^0'=2*meter, [ 3<=__const_20^0 && 2*meter==-2+__const_20^0 && meter>=1 ], cost: 2+2*meter

Computing asymptotic complexity for rule 8
   Solved the limit problem by the following transformations:
      Created initial limit problem:
      -1-2*meter+__const_20^0 (+/+!), -2+__const_20^0 (+/+!), 2+2*meter (+), 3+2*meter-__const_20^0 (+/+!) [not solved]

      removing all constraints (solved by SMT)
      resulting limit problem: [solved]

      applying transformation rule (C) using substitution {meter==n,__const_20^0==2+2*n}
      resulting limit problem:
      [solved]

   Solution:
      meter / n
      __const_20^0 / 2+2*n
   Resulting cost 2+2*n has complexity: Poly(n^1)

Found new complexity Poly(n^1).

Obtained the following overall complexity (w.r.t. the length of the input n):
   Complexity:  Poly(n^1)
   Cpx degree:  1
   Solved cost: 2+2*n
   Rule cost:   2+2*meter
   Rule guard:  [ 3<=__const_20^0 && 2*meter==-2+__const_20^0 ]

WORST_CASE(Omega(n^1),?)