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Initial Control flow graph problem:
  Start location: start
      0: eval -> eval : A'=-free+A, B'=B+free_1, [ A>=1+B && free>=0 && free_1>=1 ], cost: 1
      1: start -> eval : [], cost: 1

Eliminating 1 self-loops for location eval
  Self-Loop 0 has the metering function: -B+A, resulting in the new transition 2.
  Removing the self-loops: 0.

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


Applied simple chaining:
  Start location: start
      1: start -> [2] : A'=A, B'=A, [ A>=1+B && 0>=0 && 1>=1 ], cost: 1-B+A


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: start
      1: start -> [2] : A'=A, B'=A, [ A>=1+B && 0>=0 && 1>=1 ], cost: 1-B+A


Computing complexity for remaining 1 transitions.

  Found configuration with infinitely models for cost: 1-B+A
  and guard: A>=1+B && 0>=0 && 1>=1:
  B: Pos, A: Pos, where: A > B

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


The final runtime is determined by this resulting transition:
  Final Guard: A>=1+B && 0>=0 && 1>=1
  Final Cost:  1-B+A

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: n^1
  Complexity value: 1

WORST_CASE(Omega(n^1),?)