YES(?,O(n^1))
* Step 1: TrivialSCCs WORST_CASE(?,O(n^1))
    + 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: PolyRank WORST_CASE(?,O(n^1))
    + 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:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(f2) = x1
          p(f3) = x1
          p(f4) = x1
        
        Following rules are strictly oriented:
        [A >= 1 && B >= 1] ==>                               
           f2(A,B,C,D,E,F)   = A                             
                             > -1 + A                        
                             = f2(-1 + A,-1 + B,A,B,-2 + A,F)
        
        
        Following rules are weakly oriented:
        [B >= 1 && 0 >= A] ==>                
           f2(A,B,C,D,E,F)   = A              
                            >= A              
                             = f4(A,B,C,D,E,H)
        
        [0 >= B && 0 >= G] ==>                
           f2(A,B,C,D,E,F)   = A              
                            >= A              
                             = f4(A,G,C,D,E,H)
        
                      True ==>                
           f3(A,B,C,D,E,F)   = A              
                            >= A              
                             = f2(A,B,C,D,E,F)
        
        
* Step 3: KnowledgePropagation WORST_CASE(?,O(n^1))
    + 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] (A,1)
        Signature:
          {(f2,6);(f3,6);(f4,6)}
        Flow Graph:
          [0->{},1->{},2->{0,1,3},3->{0,1,3}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(n^1))