YES(?,O(n^1))
* Step 1: UnsatPaths WORST_CASE(?,O(n^1))
    + 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:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(3,2)]
* Step 2: TrivialSCCs WORST_CASE(?,O(n^1))
    + 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->{1,3}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 3: PolyRank WORST_CASE(?,O(n^1))
    + 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:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
            p(f1) = x1
            p(f2) = x1
          p(f300) = x1
        
        Following rules are strictly oriented:
        [A >= 1 && A + B >= 1 && B >= 1] ==>                      
                             f300(A,B,C)   = A                    
                                           > -1 + A               
                                           = f300(-1 + A,-2 + A,C)
        
        
        Following rules are weakly oriented:
                      True ==>            
                 f2(A,B,C)   = A          
                            >= A          
                             = f300(A,B,C)
        
                  [0 >= B] ==>            
               f300(A,B,C)   = A          
                            >= A          
                             = f1(A,B,D)  
        
        [B >= 1 && 0 >= A] ==>            
               f300(A,B,C)   = A          
                            >= A          
                             = f1(A,B,D)  
        
        
* Step 4: KnowledgePropagation WORST_CASE(?,O(n^1))
    + 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] (A,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3},1->{},2->{},3->{1,3}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(n^1))