YES(?,O(n^1))
* Step 1: TrivialSCCs WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. f4(A,B,C,D,E,F,G) -> f4(A + B,B,C,D,E,F,G) [-1 + -1*B >= 0 && A >= 0]     (?,1)
          1. f4(A,B,C,D,E,F,G) -> f6(A,B,0,0,0,0,0)     [-1 + -1*B >= 0 && 0 >= 1 + A] (?,1)
          2. f5(A,B,C,D,E,F,G) -> f4(A,B,C,D,E,F,G)     [0 >= 1 + B]                   (1,1)
          3. f5(A,B,C,D,E,F,G) -> f6(A,B,0,0,0,0,0)     [B >= 0]                       (1,1)
        Signature:
          {(f4,7);(f5,7);(f6,7)}
        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. f4(A,B,C,D,E,F,G) -> f4(A + B,B,C,D,E,F,G) [-1 + -1*B >= 0 && A >= 0]     (?,1)
          1. f4(A,B,C,D,E,F,G) -> f6(A,B,0,0,0,0,0)     [-1 + -1*B >= 0 && 0 >= 1 + A] (1,1)
          2. f5(A,B,C,D,E,F,G) -> f4(A,B,C,D,E,F,G)     [0 >= 1 + B]                   (1,1)
          3. f5(A,B,C,D,E,F,G) -> f6(A,B,0,0,0,0,0)     [B >= 0]                       (1,1)
        Signature:
          {(f4,7);(f5,7);(f6,7)}
        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(f4) = 1 + x1
          p(f5) = 1 + x1
          p(f6) = 1 + x1
        
        Following rules are strictly oriented:
        [-1 + -1*B >= 0 && A >= 0] ==>                      
                 f4(A,B,C,D,E,F,G)   = 1 + A                
                                     > 1 + A + B            
                                     = f4(A + B,B,C,D,E,F,G)
        
        
        Following rules are weakly oriented:
        [-1 + -1*B >= 0 && 0 >= 1 + A] ==>                  
                     f4(A,B,C,D,E,F,G)   = 1 + A            
                                        >= 1 + A            
                                         = f6(A,B,0,0,0,0,0)
        
                          [0 >= 1 + B] ==>                  
                     f5(A,B,C,D,E,F,G)   = 1 + A            
                                        >= 1 + A            
                                         = f4(A,B,C,D,E,F,G)
        
                              [B >= 0] ==>                  
                     f5(A,B,C,D,E,F,G)   = 1 + A            
                                        >= 1 + A            
                                         = f6(A,B,0,0,0,0,0)
        
        
* Step 3: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. f4(A,B,C,D,E,F,G) -> f4(A + B,B,C,D,E,F,G) [-1 + -1*B >= 0 && A >= 0]     (1 + A,1)
          1. f4(A,B,C,D,E,F,G) -> f6(A,B,0,0,0,0,0)     [-1 + -1*B >= 0 && 0 >= 1 + A] (1,1)    
          2. f5(A,B,C,D,E,F,G) -> f4(A,B,C,D,E,F,G)     [0 >= 1 + B]                   (1,1)    
          3. f5(A,B,C,D,E,F,G) -> f6(A,B,0,0,0,0,0)     [B >= 0]                       (1,1)    
        Signature:
          {(f4,7);(f5,7);(f6,7)}
        Flow Graph:
          [0->{0,1},1->{},2->{0,1},3->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(n^1))