YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. f3(A,B,C,D) -> f1(0,B,C,D)          True                                 (1,1)
          1. f1(A,B,C,D) -> f2(A,B,C,E)          [B >= C]                             (?,1)
          2. f1(A,B,C,D) -> f2(1,1 + B,C,E)      [0 >= A && 1 + B = C]                (?,1)
          3. f1(A,B,C,D) -> f1(0,1 + B,-1 + C,D) [C >= 1 + B && 0 >= A && C >= 2 + B] (?,1)
        Signature:
          {(f1,4);(f2,4);(f3,4)}
        Flow Graph:
          [0->{1,2,3},1->{},2->{},3->{1,2,3}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          f3(A,B,C,D) -> f1(0,B,C,D)           True
          f1(A,B,C,D) -> f2(A,B,C,E)           [B >= C]
          f1(A,B,C,D) -> f2(1,1 + B,C,E)       [0 >= A && 1 + B = C]
          f1(A,B,C,D) -> f1(0,1 + B,-1 + C,D)  [C >= 1 + B && 0 >= A && C >= 2 + B]
        Signature:
          {(f1,4);(f2,4);(f3,4)}
        Rule Graph:
          [0->{1,2,3},1->{},2->{},3->{1,2,3}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[3] c: [3]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          f3(A,B,C,D) -> f1(0,B,C,D)           True
          f1(A,B,C,D) -> f2(A,B,C,E)           [B >= C]
          f1(A,B,C,D) -> f2(1,1 + B,C,E)       [0 >= A && 1 + B = C]
          f1(A,B,C,D) -> f1(0,1 + B,-1 + C,D)  [C >= 1 + B && 0 >= A && C >= 2 + B]
        Signature:
          {(f1,4);(f2,4);(f3,4)}
        Rule Graph:
          [0->{1,2,3},1->{},2->{},3->{1,2,3}]
        ,We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[3] c: [3])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES