WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: l3
      0: l0 -> l1 : x^0'=x^post_1, [ x^0<=0 && x^1_1==x^0 && x^2_1==x^1_1 && x^3_1==x^2_1 && x^post_1==x^3_1 ], cost: 1
      1: l0 -> l2 : x^0'=x^post_2, [ 1<=x^0 && x^1_2==x^0 && x^post_2==-1+x^1_2 ], cost: 1
      2: l2 -> l0 : x^0'=x^post_3, [ x^post_3==x^0 ], cost: 1
      3: l3 -> l2 : x^0'=x^post_4, [ x^0==x^post_4 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      3: l3 -> l2 : x^0'=x^post_4, [ x^0==x^post_4 ], cost: 1

Removed unreachable and leaf rules:
   Start location: l3
      1: l0 -> l2 : x^0'=x^post_2, [ 1<=x^0 && x^1_2==x^0 && x^post_2==-1+x^1_2 ], cost: 1
      2: l2 -> l0 : x^0'=x^post_3, [ x^post_3==x^0 ], cost: 1
      3: l3 -> l2 : x^0'=x^post_4, [ x^0==x^post_4 ], cost: 1

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

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l3
      4: l2 -> l2 : x^0'=-1+x^0, [ 1<=x^0 ], cost: 2
      3: l3 -> l2 : [], cost: 1

Accelerating simple loops of location 2.
   Accelerating the following rules:
      4: l2 -> l2 : x^0'=-1+x^0, [ 1<=x^0 ], cost: 2

   Accelerated rule 4 with backward acceleration, yielding the new rule 5.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 4.

Accelerated all simple loops using metering functions (where possible):
   Start location: l3
      5: l2 -> l2 : x^0'=0, [ x^0>=0 ], cost: 2*x^0
      3: l3 -> l2 : [], cost: 1

Chained accelerated rules (with incoming rules):
   Start location: l3
      3: l3 -> l2 : [], cost: 1
      6: l3 -> l2 : x^0'=0, [ x^0>=0 ], cost: 1+2*x^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l3
      6: l3 -> l2 : x^0'=0, [ x^0>=0 ], cost: 1+2*x^0

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l3
      6: l3 -> l2 : x^0'=0, [ x^0>=0 ], cost: 1+2*x^0

Computing asymptotic complexity for rule 6
   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:  [ x^0==x^post_4 ]

WORST_CASE(Omega(1),?)