Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: start
      0: eval -> eval : A'=-1+A, B'=free, [ A>=0 ], cost: 1
      1: eval -> eval : B'=-1+B, [ B>=0 ], cost: 1
      2: start -> eval : [], cost: 1

Eliminating 2 self-loops for location eval
  Self-Loop 0 has the metering function: 1+A, resulting in the new transition 3.
  Self-Loop 1 has the metering function: 1+B, resulting in the new transition 4.
  Found this metering function when nesting loops: 1+A,
  and nested parallel self-loops 0 (outer loop) and 4 (inner loop), obtaining the new transitions: 5, 6.
  Removing the self-loops: 0 1.

Removed all Self-loops using metering functions (where possible):
  Start location: start
      3: eval -> [2] : A'=-1, B'=free, [ A>=0 ], cost: 1+A
      4: eval -> [2] : B'=-1, [ B>=0 ], cost: 1+B
      5: eval -> [2] : A'=-1, B'=-1, [ A>=0 && free>=0 ], cost: 2+(1+A)*free+2*A
      6: eval -> [2] : A'=-1, B'=-1, [ B>=0 && A>=0 && free>=0 ], cost: 3+B+(1+A)*free+2*A
      2: start -> eval : [], cost: 1


Applied chaining over branches and pruning:
  Start location: start
      7: start -> [2] : A'=-1, B'=free, [ A>=0 ], cost: 2+A
      8: start -> [2] : B'=-1, [ B>=0 ], cost: 2+B
      9: start -> [2] : A'=-1, B'=-1, [ A>=0 && free>=0 ], cost: 3+(1+A)*free+2*A
     10: start -> [2] : A'=-1, B'=-1, [ B>=0 && A>=0 && free>=0 ], cost: 4+B+(1+A)*free+2*A


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: start
      7: start -> [2] : A'=-1, B'=free, [ A>=0 ], cost: 2+A
      8: start -> [2] : B'=-1, [ B>=0 ], cost: 2+B
      9: start -> [2] : A'=-1, B'=-1, [ A>=0 && free>=0 ], cost: 3+(1+A)*free+2*A
     10: start -> [2] : A'=-1, B'=-1, [ B>=0 && A>=0 && free>=0 ], cost: 4+B+(1+A)*free+2*A


Computing complexity for remaining 4 transitions.

  Found configuration with infinitely models for cost: 2+A
  and guard: A>=0:
  A: Pos

Found new complexity n^1, because: Found infinity configuration.

  Found configuration with infinitely models for cost: 3+(1+A)*free+2*A
  and guard: A>=0 && free>=0:
  free: Pos, A: Pos

Found new complexity INF, because: Found infinity configuration.


The final runtime is determined by this resulting transition:
  Final Guard: A>=0 && free>=0
  Final Cost:  3+(1+A)*free+2*A

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

WORST_CASE(INF,?)