Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f20 -> f28 : [ 0>=A ], cost: 1
      6: f20 -> f28 : A'=0, B'=free_2, G'=0, H'=free_2, Q'=free_2, [ A>=1 ], cost: 1
      3: f28 -> f40 : [ A>=1 ], cost: 1
      4: f28 -> f40 : C'=0, D'=free, E'=free, F'=free, [ 0>=A && 999+B>=free ], cost: 1
      5: f28 -> f40 : A'=1, C'=0, D'=free_1, E'=free_1, F'=free_1, [ 0>=A && free_1>=1000+B ], cost: 1
      1: f40 -> f40 : [], cost: 1
      2: f42 -> f45 : [], cost: 1
      7: f0 -> f20 : A'=1, J'=free_3, [ 0>=free_3 ], cost: 1
      8: f0 -> f20 : A'=1, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 && free_5>=1 ], cost: 1
      9: f0 -> f40 : A'=1, J'=0, K'=1, L'=free_6, M'=free_6, N'=free_6, O'=free_6, [ 0>=free_6 && free_7>=1 ], cost: 1


Simplified the transitions:
  Start location: f0
      0: f20 -> f28 : [ 0>=A ], cost: 1
      6: f20 -> f28 : A'=0, B'=free_2, G'=0, H'=free_2, Q'=free_2, [ A>=1 ], cost: 1
      3: f28 -> f40 : [ A>=1 ], cost: 1
      4: f28 -> f40 : C'=0, D'=free, E'=free, F'=free, [ 0>=A && 999+B>=free ], cost: 1
      5: f28 -> f40 : A'=1, C'=0, D'=free_1, E'=free_1, F'=free_1, [ 0>=A && free_1>=1000+B ], cost: 1
      1: f40 -> f40 : [], cost: 1
      7: f0 -> f20 : A'=1, J'=free_3, [ 0>=free_3 ], cost: 1
      8: f0 -> f20 : A'=1, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 ], cost: 1
      9: f0 -> f40 : A'=1, J'=0, K'=1, L'=free_6, M'=free_6, N'=free_6, O'=free_6, [ 0>=free_6 ], cost: 1

Eliminating 1 self-loops for location f40
  Self-Loop 1 has unbounded runtime, resulting in the new transition 10.
  Removing the self-loops: 1.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f20 -> f28 : [ 0>=A ], cost: 1
      6: f20 -> f28 : A'=0, B'=free_2, G'=0, H'=free_2, Q'=free_2, [ A>=1 ], cost: 1
      3: f28 -> f40 : [ A>=1 ], cost: 1
      4: f28 -> f40 : C'=0, D'=free, E'=free, F'=free, [ 0>=A && 999+B>=free ], cost: 1
      5: f28 -> f40 : A'=1, C'=0, D'=free_1, E'=free_1, F'=free_1, [ 0>=A && free_1>=1000+B ], cost: 1
     10: f40 -> [6] : [], cost: INF
      7: f0 -> f20 : A'=1, J'=free_3, [ 0>=free_3 ], cost: 1
      8: f0 -> f20 : A'=1, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 ], cost: 1
      9: f0 -> f40 : A'=1, J'=0, K'=1, L'=free_6, M'=free_6, N'=free_6, O'=free_6, [ 0>=free_6 ], cost: 1


Applied chaining over branches and pruning:
  Start location: f0
      3: f28 -> f40 : [ A>=1 ], cost: 1
      4: f28 -> f40 : C'=0, D'=free, E'=free, F'=free, [ 0>=A && 999+B>=free ], cost: 1
      5: f28 -> f40 : A'=1, C'=0, D'=free_1, E'=free_1, F'=free_1, [ 0>=A && free_1>=1000+B ], cost: 1
     10: f40 -> [6] : [], cost: INF
     11: f0 -> f28 : A'=0, B'=free_2, G'=0, H'=free_2, Q'=free_2, J'=free_3, [ 0>=free_3 && 1>=1 ], cost: 2
     12: f0 -> f28 : A'=0, B'=free_2, G'=0, H'=free_2, Q'=free_2, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 && 1>=1 ], cost: 2
      9: f0 -> f40 : A'=1, J'=0, K'=1, L'=free_6, M'=free_6, N'=free_6, O'=free_6, [ 0>=free_6 ], cost: 1


Applied chaining over branches and pruning:
  Start location: f0
     10: f40 -> [6] : [], cost: INF
      9: f0 -> f40 : A'=1, J'=0, K'=1, L'=free_6, M'=free_6, N'=free_6, O'=free_6, [ 0>=free_6 ], cost: 1
     13: f0 -> f40 : A'=0, B'=free_2, C'=0, D'=free, E'=free, F'=free, G'=0, H'=free_2, Q'=free_2, J'=free_3, [ 0>=free_3 && 1>=1 && 0>=0 && 999+free_2>=free ], cost: 3
     14: f0 -> f40 : A'=1, B'=free_2, C'=0, D'=free_1, E'=free_1, F'=free_1, G'=0, H'=free_2, Q'=free_2, J'=free_3, [ 0>=free_3 && 1>=1 && 0>=0 && free_1>=1000+free_2 ], cost: 3
     15: f0 -> f40 : A'=0, B'=free_2, C'=0, D'=free, E'=free, F'=free, G'=0, H'=free_2, Q'=free_2, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 && 1>=1 && 0>=0 && 999+free_2>=free ], cost: 3
     16: f0 -> f40 : A'=1, B'=free_2, C'=0, D'=free_1, E'=free_1, F'=free_1, G'=0, H'=free_2, Q'=free_2, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 && 1>=1 && 0>=0 && free_1>=1000+free_2 ], cost: 3


Applied chaining over branches and pruning:
  Start location: f0
     17: f0 -> [6] : A'=1, J'=0, K'=1, L'=free_6, M'=free_6, N'=free_6, O'=free_6, [ 0>=free_6 ], cost: INF
     18: f0 -> [6] : A'=0, B'=free_2, C'=0, D'=free, E'=free, F'=free, G'=0, H'=free_2, Q'=free_2, J'=free_3, [ 0>=free_3 && 1>=1 && 0>=0 && 999+free_2>=free ], cost: INF
     19: f0 -> [6] : A'=1, B'=free_2, C'=0, D'=free_1, E'=free_1, F'=free_1, G'=0, H'=free_2, Q'=free_2, J'=free_3, [ 0>=free_3 && 1>=1 && 0>=0 && free_1>=1000+free_2 ], cost: INF
     20: f0 -> [6] : A'=0, B'=free_2, C'=0, D'=free, E'=free, F'=free, G'=0, H'=free_2, Q'=free_2, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 && 1>=1 && 0>=0 && 999+free_2>=free ], cost: INF
     21: f0 -> [6] : A'=1, B'=free_2, C'=0, D'=free_1, E'=free_1, F'=free_1, G'=0, H'=free_2, Q'=free_2, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 && 1>=1 && 0>=0 && free_1>=1000+free_2 ], cost: INF


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
     17: f0 -> [6] : A'=1, J'=0, K'=1, L'=free_6, M'=free_6, N'=free_6, O'=free_6, [ 0>=free_6 ], cost: INF
     18: f0 -> [6] : A'=0, B'=free_2, C'=0, D'=free, E'=free, F'=free, G'=0, H'=free_2, Q'=free_2, J'=free_3, [ 0>=free_3 && 1>=1 && 0>=0 && 999+free_2>=free ], cost: INF
     19: f0 -> [6] : A'=1, B'=free_2, C'=0, D'=free_1, E'=free_1, F'=free_1, G'=0, H'=free_2, Q'=free_2, J'=free_3, [ 0>=free_3 && 1>=1 && 0>=0 && free_1>=1000+free_2 ], cost: INF
     20: f0 -> [6] : A'=0, B'=free_2, C'=0, D'=free, E'=free, F'=free, G'=0, H'=free_2, Q'=free_2, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 && 1>=1 && 0>=0 && 999+free_2>=free ], cost: INF
     21: f0 -> [6] : A'=1, B'=free_2, C'=0, D'=free_1, E'=free_1, F'=free_1, G'=0, H'=free_2, Q'=free_2, J'=0, K'=1, L'=free_4, M'=free_4, N'=free_4, O'=free_4, [ free_4>=1 && 1>=1 && 0>=0 && free_1>=1000+free_2 ], cost: INF


Computing complexity for remaining 5 transitions.

Found new complexity INF, because: INF sat.


The final runtime is determined by this resulting transition:
  Final Guard: 0>=free_6
  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,?)