Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f2
      0: f1 -> f1 : C'=free_2, D'=free_1, E'=free, [ free_1>=1 && A>=1+B ], cost: 1
      1: f1 -> f1 : C'=free_5, D'=free_4, E'=free_3, [ 0>=1+free_4 && A>=1+B ], cost: 1
      2: f1 -> f1 : C'=free_6, D'=0, [ A>=1+B ], cost: 1
      3: f1 -> f300 : C'=free_8, F'=free_7, [ B>=A ], cost: 1
      4: f2 -> f1 : G'=free_9, H'=free_9, [], cost: 1


Simplified the transitions:
  Start location: f2
      0: f1 -> f1 : C'=free_2, D'=free_1, E'=free, [ free_1>=1 && A>=1+B ], cost: 1
      1: f1 -> f1 : C'=free_5, D'=free_4, E'=free_3, [ 0>=1+free_4 && A>=1+B ], cost: 1
      2: f1 -> f1 : C'=free_6, D'=0, [ A>=1+B ], cost: 1
      4: f2 -> f1 : G'=free_9, H'=free_9, [], cost: 1

Eliminating 3 self-loops for location f1
  Self-Loop 0 has unbounded runtime, resulting in the new transition 5.
  Self-Loop 1 has unbounded runtime, resulting in the new transition 6.
  Self-Loop 2 has unbounded runtime, resulting in the new transition 7.
  Removing the self-loops: 0 1 2.

Removed all Self-loops using metering functions (where possible):
  Start location: f2
      5: f1 -> [3] : [ free_1>=1 && A>=1+B ], cost: INF
      6: f1 -> [3] : [ 0>=1+free_4 && A>=1+B ], cost: INF
      7: f1 -> [3] : [ A>=1+B ], cost: INF
      4: f2 -> f1 : G'=free_9, H'=free_9, [], cost: 1


Applied chaining over branches and pruning:
  Start location: f2
      8: f2 -> [3] : G'=free_9, H'=free_9, [ free_1>=1 && A>=1+B ], cost: INF
      9: f2 -> [3] : G'=free_9, H'=free_9, [ 0>=1+free_4 && A>=1+B ], cost: INF
     10: f2 -> [3] : G'=free_9, H'=free_9, [ A>=1+B ], cost: INF


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f2
      8: f2 -> [3] : G'=free_9, H'=free_9, [ free_1>=1 && A>=1+B ], cost: INF
      9: f2 -> [3] : G'=free_9, H'=free_9, [ 0>=1+free_4 && A>=1+B ], cost: INF
     10: f2 -> [3] : G'=free_9, H'=free_9, [ A>=1+B ], 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: free_1>=1 && A>=1+B
  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,?)