YES(?,O(n^1))
* Step 1: TrivialSCCs WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. f5(A,B)   -> f5(-1 + A,B) [A >= 1] (?,1)
          1. f5(A,B)   -> f1(A,C)      [0 >= A] (?,1)
          2. f300(A,B) -> f5(-1 + A,B) [A >= 1] (1,1)
          3. f300(A,B) -> f1(A,C)      [0 >= A] (1,1)
        Signature:
          {(f1,2);(f300,2);(f5,2)}
        Flow Graph:
          [0->{0,1},1->{},2->{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. f5(A,B)   -> f5(-1 + A,B) [A >= 1] (?,1)
          1. f5(A,B)   -> f1(A,C)      [0 >= A] (1,1)
          2. f300(A,B) -> f5(-1 + A,B) [A >= 1] (1,1)
          3. f300(A,B) -> f1(A,C)      [0 >= A] (1,1)
        Signature:
          {(f1,2);(f300,2);(f5,2)}
        Flow Graph:
          [0->{0,1},1->{},2->{0,1},3->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
            p(f1) = x1
          p(f300) = x1
            p(f5) = x1
        
        Following rules are strictly oriented:
           [A >= 1] ==>             
            f5(A,B)   = A           
                      > -1 + A      
                      = f5(-1 + A,B)
        
           [A >= 1] ==>             
          f300(A,B)   = A           
                      > -1 + A      
                      = f5(-1 + A,B)
        
        
        Following rules are weakly oriented:
           [0 >= A] ==>        
            f5(A,B)   = A      
                     >= A      
                      = f1(A,C)
        
           [0 >= A] ==>        
          f300(A,B)   = A      
                     >= A      
                      = f1(A,C)
        
        
* Step 3: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. f5(A,B)   -> f5(-1 + A,B) [A >= 1] (A,1)
          1. f5(A,B)   -> f1(A,C)      [0 >= A] (1,1)
          2. f300(A,B) -> f5(-1 + A,B) [A >= 1] (1,1)
          3. f300(A,B) -> f1(A,C)      [0 >= A] (1,1)
        Signature:
          {(f1,2);(f300,2);(f5,2)}
        Flow Graph:
          [0->{0,1},1->{},2->{0,1},3->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(n^1))