YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0. f2(A,B,C,D,E,F) -> f4(A,B,C,D,E,H)                [B >= 1 && 0 >= A] (?,1)
          1. f2(A,B,C,D,E,F) -> f4(A,G,C,D,E,H)                [0 >= B && 0 >= G] (?,1)
          2. f3(A,B,C,D,E,F) -> f2(A,B,C,D,E,F)                True               (1,1)
          3. f2(A,B,C,D,E,F) -> f2(-1 + A,-1 + B,A,B,-2 + A,F) [A >= 1 && B >= 1] (?,1)
        Signature:
          {(f2,6);(f3,6);(f4,6)}
        Flow Graph:
          [0->{},1->{},2->{0,1,3},3->{0,1,3}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: Looptree YES
    + Considered Problem:
        Rules:
          0. f2(A,B,C,D,E,F) -> f4(A,B,C,D,E,H)                [B >= 1 && 0 >= A] (1,1)
          1. f2(A,B,C,D,E,F) -> f4(A,G,C,D,E,H)                [0 >= B && 0 >= G] (1,1)
          2. f3(A,B,C,D,E,F) -> f2(A,B,C,D,E,F)                True               (1,1)
          3. f2(A,B,C,D,E,F) -> f2(-1 + A,-1 + B,A,B,-2 + A,F) [A >= 1 && B >= 1] (?,1)
        Signature:
          {(f2,6);(f3,6);(f4,6)}
        Flow Graph:
          [0->{},1->{},2->{0,1,3},3->{0,1,3}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[3] c: [3]

YES