YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. f0(A) -> f5(B)     [B >= 1]                 (1,1)
          1. f5(A) -> f5(1 + A) [-1 + A >= 0 && 19 >= A] (?,1)
          2. f0(A) -> f12(B)    [0 >= B]                 (1,1)
          3. f5(A) -> f12(A)    [-1 + A >= 0 && A >= 20] (?,1)
        Signature:
          {(f0,1);(f12,1);(f5,1)}
        Flow Graph:
          [0->{1,3},1->{1,3},2->{},3->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          f0(A) -> f5(B)      [B >= 1]
          f5(A) -> f5(1 + A)  [-1 + A >= 0 && 19 >= A]
          f0(A) -> f12(B)     [0 >= B]
          f5(A) -> f12(A)     [-1 + A >= 0 && A >= 20]
        Signature:
          {(f0,1);(f12,1);(f5,1)}
        Rule Graph:
          [0->{1,3},1->{1,3},2->{},3->{}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[1] c: [1]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          f0(A) -> f5(B)      [B >= 1]
          f5(A) -> f5(1 + A)  [-1 + A >= 0 && 19 >= A]
          f0(A) -> f12(B)     [0 >= B]
          f5(A) -> f12(A)     [-1 + A >= 0 && A >= 20]
        Signature:
          {(f0,1);(f12,1);(f5,1)}
        Rule Graph:
          [0->{1,3},1->{1,3},2->{},3->{}]
        ,We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[1] c: [1])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES