Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f9 : A'=free, B'=free, C'=free, D'=0, [], cost: 1
      1: f9 -> f9 : D'=1+D, E'=free_1, [ C>=1+D ], cost: 1
      8: f9 -> f24 : [ D>=C ], cost: 1
      2: f24 -> f24 : [ 0>=1+free_2 ], cost: 1
      3: f24 -> f24 : [], cost: 1
      7: f24 -> f37 : [], cost: 1
      4: f37 -> f37 : [ free_3>=1+free_4 ], cost: 1
      5: f37 -> f37 : [], cost: 1
      6: f37 -> f48 : [], cost: 1

Removing duplicate transition: 2.
Removing duplicate transition: 4.

Simplified the transitions:
  Start location: f0
      0: f0 -> f9 : A'=free, B'=free, C'=free, D'=0, [], cost: 1
      1: f9 -> f9 : D'=1+D, E'=free_1, [ C>=1+D ], cost: 1
      8: f9 -> f24 : [ D>=C ], cost: 1
      3: f24 -> f24 : [], cost: 1
      7: f24 -> f37 : [], cost: 1
      5: f37 -> f37 : [], cost: 1

Eliminating 1 self-loops for location f9
  Self-Loop 1 has the metering function: C-D, resulting in the new transition 9.
  Removing the self-loops: 1.
Eliminating 1 self-loops for location f24
  Self-Loop 3 has unbounded runtime, resulting in the new transition 10.
  Removing the self-loops: 3.
Eliminating 1 self-loops for location f37
  Self-Loop 5 has unbounded runtime, resulting in the new transition 11.
  Removing the self-loops: 5.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f0 -> f9 : A'=free, B'=free, C'=free, D'=0, [], cost: 1
      9: f9 -> [5] : D'=C, E'=free_1, [ C>=1+D ], cost: C-D
     10: f24 -> [6] : [], cost: INF
     11: f37 -> [7] : [], cost: INF
      8: [5] -> f24 : [ D>=C ], cost: 1
      7: [6] -> f37 : [], cost: 1


Applied simple chaining:
  Start location: f0
      0: f0 -> [7] : A'=free, B'=free, C'=free, D'=free, E'=free_1, [ free>=1 && free>=free ], cost: INF


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      0: f0 -> [7] : A'=free, B'=free, C'=free, D'=free, E'=free_1, [ free>=1 && free>=free ], 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: free>=1 && free>=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,?)