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Initial Control flow graph problem:
  Start location: f1
      1: f2 -> f2 : B'=free_3, C'=free_4, E'=free_5, [ A>=1 ], cost: 1
      0: f2 -> f300 : B'=free, C'=free_1, D'=free_2, [ 0>=A ], cost: 1
      2: f1 -> f2 : F'=free_6, G'=free_7, H'=free_8, Q'=free_8, J'=free_6, [], cost: 1


Simplified the transitions:
  Start location: f1
      1: f2 -> f2 : B'=free_3, C'=free_4, E'=free_5, [ A>=1 ], cost: 1
      2: f1 -> f2 : F'=free_6, G'=free_7, H'=free_8, Q'=free_8, J'=free_6, [], cost: 1

Eliminating 1 self-loops for location f2
  Self-Loop 1 has unbounded runtime, resulting in the new transition 3.
  Removing the self-loops: 1.

Removed all Self-loops using metering functions (where possible):
  Start location: f1
      3: f2 -> [3] : [ A>=1 ], cost: INF
      2: f1 -> f2 : F'=free_6, G'=free_7, H'=free_8, Q'=free_8, J'=free_6, [], cost: 1


Applied simple chaining:
  Start location: f1
      2: f1 -> [3] : F'=free_6, G'=free_7, H'=free_8, Q'=free_8, J'=free_6, [ A>=1 ], cost: INF


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f1
      2: f1 -> [3] : F'=free_6, G'=free_7, H'=free_8, Q'=free_8, J'=free_6, [ A>=1 ], cost: INF


Computing complexity for remaining 1 transitions.

Found new complexity INF, because: INF sat.


The final runtime is determined by this resulting transition:
  Final Guard: A>=1
  Final Cost:  INF

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

WORST_CASE(INF,?)