YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0. f2(A,B,C)   -> f300(A,B,C)           True                             (1,1)
          1. f300(A,B,C) -> f1(A,B,D)             [0 >= B]                         (?,1)
          2. f300(A,B,C) -> f1(A,B,D)             [B >= 1 && 0 >= A]               (?,1)
          3. f300(A,B,C) -> f300(-1 + A,-2 + A,C) [A >= 1 && A + B >= 1 && B >= 1] (?,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3},1->{},2->{},3->{1,2,3}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f2(A,B,C)   -> f300(A,B,C)           True                             (1,1)
          1. f300(A,B,C) -> f1(A,B,D)             [0 >= B]                         (1,1)
          2. f300(A,B,C) -> f1(A,B,D)             [B >= 1 && 0 >= A]               (1,1)
          3. f300(A,B,C) -> f300(-1 + A,-2 + A,C) [A >= 1 && A + B >= 1 && B >= 1] (?,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3},1->{},2->{},3->{1,2,3}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(3,2)]
* Step 3: Looptree YES
    + Considered Problem:
        Rules:
          0. f2(A,B,C)   -> f300(A,B,C)           True                             (1,1)
          1. f300(A,B,C) -> f1(A,B,D)             [0 >= B]                         (1,1)
          2. f300(A,B,C) -> f1(A,B,D)             [B >= 1 && 0 >= A]               (1,1)
          3. f300(A,B,C) -> f300(-1 + A,-2 + A,C) [A >= 1 && A + B >= 1 && B >= 1] (?,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3},1->{},2->{},3->{1,3}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[3] c: [3]

YES