MAYBE
* Step 1: ArgumentFilter MAYBE
    + Considered Problem:
        Rules:
          0. f3(A,B) -> f12(A,B)    [A >= 10]          (?,1)
          1. f3(A,B) -> f3(1 + A,A) [9 >= A && A >= 6] (?,1)
          2. f3(A,B) -> f3(1 + A,A) [9 >= A && 4 >= A] (?,1)
          3. f3(A,B) -> f3(1 + C,B) [A = 5]            (?,1)
          4. f0(A,B) -> f3(0,B)     True               (1,1)
        Signature:
          {(f0,2);(f12,2);(f3,2)}
        Flow Graph:
          [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}]
        
    + Applied Processor:
        ArgumentFilter [1]
    + Details:
        We remove following argument positions: [1].
* Step 2: UnsatPaths MAYBE
    + Considered Problem:
        Rules:
          0. f3(A) -> f12(A)    [A >= 10]          (?,1)
          1. f3(A) -> f3(1 + A) [9 >= A && A >= 6] (?,1)
          2. f3(A) -> f3(1 + A) [9 >= A && 4 >= A] (?,1)
          3. f3(A) -> f3(1 + C) [A = 5]            (?,1)
          4. f0(A) -> f3(0)     True               (1,1)
        Signature:
          {(f0,2);(f12,2);(f3,2)}
        Flow Graph:
          [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(1,2),(1,3),(2,0),(2,1),(4,0),(4,1),(4,3)]
* Step 3: FromIts MAYBE
    + Considered Problem:
        Rules:
          0. f3(A) -> f12(A)    [A >= 10]          (?,1)
          1. f3(A) -> f3(1 + A) [9 >= A && A >= 6] (?,1)
          2. f3(A) -> f3(1 + A) [9 >= A && 4 >= A] (?,1)
          3. f3(A) -> f3(1 + C) [A = 5]            (?,1)
          4. f0(A) -> f3(0)     True               (1,1)
        Signature:
          {(f0,2);(f12,2);(f3,2)}
        Flow Graph:
          [0->{},1->{0,1},2->{2,3},3->{0,1,2,3},4->{2}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 4: AddSinks MAYBE
    + Considered Problem:
        Rules:
          f3(A) -> f12(A)     [A >= 10]
          f3(A) -> f3(1 + A)  [9 >= A && A >= 6]
          f3(A) -> f3(1 + A)  [9 >= A && 4 >= A]
          f3(A) -> f3(1 + C)  [A = 5]
          f0(A) -> f3(0)      True
        Signature:
          {(f0,2);(f12,2);(f3,2)}
        Rule Graph:
          [0->{},1->{0,1},2->{2,3},3->{0,1,2,3},4->{2}]
    + Applied Processor:
        AddSinks
    + Details:
        ()
* Step 5: Failure MAYBE
  + Considered Problem:
      Rules:
        f3(A)  -> f12(A)        [A >= 10]
        f3(A)  -> f3(1 + A)     [9 >= A && A >= 6]
        f3(A)  -> f3(1 + A)     [9 >= A && 4 >= A]
        f3(A)  -> f3(1 + C)     [A = 5]
        f0(A)  -> f3(0)         True
        f12(A) -> exitus616(A)  True
      Signature:
        {(exitus616,1);(f0,2);(f12,2);(f3,2)}
      Rule Graph:
        [0->{5},1->{0,1},2->{2,3},3->{0,1,2,3},4->{2}]
  + Applied Processor:
      Decompose Greedy
  + Details:
      We construct a looptree:
        P: [0,1,2,3,4,5]
        |
        +- p:[2,3] c: []
        |
        `- p:[1] c: [1]
MAYBE