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Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f5 : A'=free, [ free>=1 ], cost: 1
      2: f0 -> f12 : A'=free_1, [ 0>=free_1 ], cost: 1
      1: f5 -> f5 : A'=1+A, [ 19>=A ], cost: 1
      3: f5 -> f12 : [ A>=20 ], cost: 1


Simplified the transitions:
  Start location: f0
      0: f0 -> f5 : A'=free, [ free>=1 ], cost: 1
      1: f5 -> f5 : A'=1+A, [ 19>=A ], cost: 1

Eliminating 1 self-loops for location f5
  Self-Loop 1 has the metering function: 20-A, resulting in the new transition 4.
  Removing the self-loops: 1.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f0 -> f5 : A'=free, [ free>=1 ], cost: 1
      4: f5 -> [3] : A'=20, [ 19>=A ], cost: 20-A


Applied simple chaining:
  Start location: f0
      0: f0 -> [3] : A'=20, [ free>=1 && 19>=free ], cost: 21-free


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      0: f0 -> [3] : A'=20, [ free>=1 && 19>=free ], cost: 21-free


Computing complexity for remaining 1 transitions.


The final runtime is determined by this resulting transition:
  Final Guard: 
  Final Cost:  1

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: const
  Complexity value: 0

WORST_CASE(Omega(1),?)