YES(?,O(n^1))
* Step 1: TrivialSCCs WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. f2(A,B,C) -> f2(A + -1*B,1 + B,C) [-1 + B >= 0 && A >= 1] (?,1)
          1. f3(A,B,C) -> f2(A,B,C)            [B >= 1]                (1,1)
          2. f2(A,B,C) -> f4(A,B,D)            [-1 + B >= 0 && 0 >= A] (?,1)
          3. f3(A,B,C) -> f4(A,B,D)            [0 >= B]                (1,1)
        Signature:
          {(f2,3);(f3,3);(f4,3)}
        Flow Graph:
          [0->{0,2},1->{0,2},2->{},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) -> f2(A + -1*B,1 + B,C) [-1 + B >= 0 && A >= 1] (?,1)
          1. f3(A,B,C) -> f2(A,B,C)            [B >= 1]                (1,1)
          2. f2(A,B,C) -> f4(A,B,D)            [-1 + B >= 0 && 0 >= A] (1,1)
          3. f3(A,B,C) -> f4(A,B,D)            [0 >= B]                (1,1)
        Signature:
          {(f2,3);(f3,3);(f4,3)}
        Flow Graph:
          [0->{0,2},1->{0,2},2->{},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:
        [-1 + B >= 0 && A >= 1] ==>                     
                      f2(A,B,C)   = A                   
                                  > A + -1*B            
                                  = f2(A + -1*B,1 + B,C)
        
        
        Following rules are weakly oriented:
                       [B >= 1] ==>          
                      f3(A,B,C)   = A        
                                 >= A        
                                  = f2(A,B,C)
        
        [-1 + B >= 0 && 0 >= A] ==>          
                      f2(A,B,C)   = A        
                                 >= A        
                                  = f4(A,B,D)
        
                       [0 >= B] ==>          
                      f3(A,B,C)   = A        
                                 >= A        
                                  = f4(A,B,D)
        
        
* Step 3: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. f2(A,B,C) -> f2(A + -1*B,1 + B,C) [-1 + B >= 0 && A >= 1] (A,1)
          1. f3(A,B,C) -> f2(A,B,C)            [B >= 1]                (1,1)
          2. f2(A,B,C) -> f4(A,B,D)            [-1 + B >= 0 && 0 >= A] (1,1)
          3. f3(A,B,C) -> f4(A,B,D)            [0 >= B]                (1,1)
        Signature:
          {(f2,3);(f3,3);(f4,3)}
        Flow Graph:
          [0->{0,2},1->{0,2},2->{},3->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(n^1))