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Initial Control flow graph problem:
  Start location: f
      0: f -> g : B'=1, C'=1, [], cost: 1
      1: g -> g : A'=-1+A, B'=2*B, [ A>0 ], cost: 1
      2: g -> h : [ A<=0 ], cost: 1
      3: h -> h : B'=-1+B, C'=2*C, [ B>0 ], cost: 1
      4: h -> i : [ B<=0 ], cost: 1
      5: i -> i : C'=-1+C, [ C>0 ], cost: 1

Eliminating 1 self-loops for location g
  Self-Loop 1 has the metering function: A, resulting in the new transition 6.
  Removing the self-loops: 1.
Eliminating 1 self-loops for location h
  Self-Loop 3 has the metering function: B, resulting in the new transition 7.
  Removing the self-loops: 3.
Eliminating 1 self-loops for location i
  Self-Loop 5 has the metering function: C, resulting in the new transition 8.
  Removing the self-loops: 5.

Removed all Self-loops using metering functions (where possible):
  Start location: f
      0: f -> g : B'=1, C'=1, [], cost: 1
      6: g -> [4] : A'=0, B'=B*2^A, [ A>0 ], cost: A
      7: h -> [5] : B'=0, C'=2^B*C, [ B>0 ], cost: B
      8: i -> [6] : C'=0, [ C>0 ], cost: C
      2: [4] -> h : [ A<=0 ], cost: 1
      4: [5] -> i : [ B<=0 ], cost: 1


Applied simple chaining:
  Start location: f
      0: f -> [6] : A'=0, B'=0, C'=0, [ A>0 && 0<=0 && 2^A>0 && 0<=0 && 2^(2^A)>0 ], cost: 3+2^A+2^(2^A)+A


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f
      0: f -> [6] : A'=0, B'=0, C'=0, [ A>0 && 0<=0 && 2^A>0 && 0<=0 && 2^(2^A)>0 ], cost: 3+2^A+2^(2^A)+A


Computing complexity for remaining 1 transitions.

  Found configuration with infinitely models for cost: 3+2^A+2^(2^A)+A
  and guard: A>0 && 0<=0 && 2^A>0 && 0<=0 && 2^(2^A)>0:
  A: Pos

Found new complexity EXP NESTED, because: Found infinity configuration.


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

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

WORST_CASE(EXP,?)