Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f14 : A'=free, B'=free, C'=0, [ 0>=free ], cost: 1
      1: f0 -> f14 : A'=free_1, B'=free_1, C'=0, [ free_1>=1024 ], cost: 1
      2: f0 -> f14 : A'=free_2, B'=free_2, C'=0, D'=free_3, [ 1023>=free_2 && free_2>=1 ], cost: 1
      3: f14 -> f14 : C'=1+C, [ free_4>=1+C ], cost: 1
      4: f14 -> f22 : [ C>=free_5 ], cost: 1


Simplified the transitions:
  Start location: f0
      0: f0 -> f14 : A'=free, B'=free, C'=0, [ 0>=free ], cost: 1
      1: f0 -> f14 : A'=free_1, B'=free_1, C'=0, [ free_1>=1024 ], cost: 1
      2: f0 -> f14 : A'=free_2, B'=free_2, C'=0, D'=free_3, [ 1023>=free_2 && free_2>=1 ], cost: 1
      3: f14 -> f14 : C'=1+C, [], cost: 1

Eliminating 1 self-loops for location f14
  Self-Loop 3 has unbounded runtime, resulting in the new transition 5.
  Removing the self-loops: 3.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f0 -> f14 : A'=free, B'=free, C'=0, [ 0>=free ], cost: 1
      1: f0 -> f14 : A'=free_1, B'=free_1, C'=0, [ free_1>=1024 ], cost: 1
      2: f0 -> f14 : A'=free_2, B'=free_2, C'=0, D'=free_3, [ 1023>=free_2 && free_2>=1 ], cost: 1
      5: f14 -> [3] : [], cost: INF


Applied chaining over branches and pruning:
  Start location: f0
      6: f0 -> [3] : A'=free, B'=free, C'=0, [ 0>=free ], cost: INF
      7: f0 -> [3] : A'=free_1, B'=free_1, C'=0, [ free_1>=1024 ], cost: INF
      8: f0 -> [3] : A'=free_2, B'=free_2, C'=0, D'=free_3, [ 1023>=free_2 && free_2>=1 ], cost: INF


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      6: f0 -> [3] : A'=free, B'=free, C'=0, [ 0>=free ], cost: INF
      7: f0 -> [3] : A'=free_1, B'=free_1, C'=0, [ free_1>=1024 ], cost: INF
      8: f0 -> [3] : A'=free_2, B'=free_2, C'=0, D'=free_3, [ 1023>=free_2 && free_2>=1 ], cost: INF


Computing complexity for remaining 3 transitions.

Found new complexity INF, because: INF sat.


The final runtime is determined by this resulting transition:
  Final Guard: 0>=free
  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,?)