YES(?,O(n^1))
* Step 1: TrivialSCCs WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B,C,D,E,F)  -> stop(A,B,C,D,E,F)                [0 >= A && B = A && C = D && E = F]                                            (?,1)
          1. start(A,B,C,D,E,F)  -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [A >= 1 && B = A && C = D && E = F]                                            (?,1)
          2. lbl71(A,B,C,D,E,F)  -> stop(A,B,C,D,E,F)                [D >= 1 + C && B = 0 && C + E = D + F && A + C = D]                            (?,1)
          3. lbl71(A,B,C,D,E,F)  -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [A + C >= 1 + D && D >= 1 + C && A + C >= D && C + E = D + F && B + D = A + C] (?,1)
          4. start0(A,B,C,D,E,F) -> start(A,A,D,D,F,F)               True                                                                           (1,1)
        Signature:
          {(lbl71,6);(start,6);(start0,6);(stop,6)}
        Flow Graph:
          [0->{},1->{2,3},2->{},3->{2,3},4->{0,1}]
        
    + 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. start(A,B,C,D,E,F)  -> stop(A,B,C,D,E,F)                [0 >= A && B = A && C = D && E = F]                                            (1,1)
          1. start(A,B,C,D,E,F)  -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [A >= 1 && B = A && C = D && E = F]                                            (1,1)
          2. lbl71(A,B,C,D,E,F)  -> stop(A,B,C,D,E,F)                [D >= 1 + C && B = 0 && C + E = D + F && A + C = D]                            (1,1)
          3. lbl71(A,B,C,D,E,F)  -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [A + C >= 1 + D && D >= 1 + C && A + C >= D && C + E = D + F && B + D = A + C] (?,1)
          4. start0(A,B,C,D,E,F) -> start(A,A,D,D,F,F)               True                                                                           (1,1)
        Signature:
          {(lbl71,6);(start,6);(start0,6);(stop,6)}
        Flow Graph:
          [0->{},1->{2,3},2->{},3->{2,3},4->{0,1}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(lbl71) = 1 + x1 + x3 + -1*x4
           p(start) = x1 + x3 + -1*x4    
          p(start0) = x1                 
            p(stop) = x1 + x3 + -1*x4    
        
        Following rules are strictly oriented:
                                   [D >= 1 + C && B = 0 && C + E = D + F && A + C = D] ==>                                 
                                                                    lbl71(A,B,C,D,E,F)   = 1 + A + C + -1*D                
                                                                                         > A + C + -1*D                    
                                                                                         = stop(A,B,C,D,E,F)               
        
        [A + C >= 1 + D && D >= 1 + C && A + C >= D && C + E = D + F && B + D = A + C] ==>                                 
                                                                    lbl71(A,B,C,D,E,F)   = 1 + A + C + -1*D                
                                                                                         > A + C + -1*D                    
                                                                                         = lbl71(A,-1 + B,-1 + C,D,1 + E,F)
        
        
        Following rules are weakly oriented:
        [0 >= A && B = A && C = D && E = F] ==>                                 
                         start(A,B,C,D,E,F)   = A + C + -1*D                    
                                             >= A + C + -1*D                    
                                              = stop(A,B,C,D,E,F)               
        
        [A >= 1 && B = A && C = D && E = F] ==>                                 
                         start(A,B,C,D,E,F)   = A + C + -1*D                    
                                             >= A + C + -1*D                    
                                              = lbl71(A,-1 + B,-1 + C,D,1 + E,F)
        
                                       True ==>                                 
                        start0(A,B,C,D,E,F)   = A                               
                                             >= A                               
                                              = start(A,A,D,D,F,F)              
        
        
* Step 3: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B,C,D,E,F)  -> stop(A,B,C,D,E,F)                [0 >= A && B = A && C = D && E = F]                                            (1,1)
          1. start(A,B,C,D,E,F)  -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [A >= 1 && B = A && C = D && E = F]                                            (1,1)
          2. lbl71(A,B,C,D,E,F)  -> stop(A,B,C,D,E,F)                [D >= 1 + C && B = 0 && C + E = D + F && A + C = D]                            (1,1)
          3. lbl71(A,B,C,D,E,F)  -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [A + C >= 1 + D && D >= 1 + C && A + C >= D && C + E = D + F && B + D = A + C] (A,1)
          4. start0(A,B,C,D,E,F) -> start(A,A,D,D,F,F)               True                                                                           (1,1)
        Signature:
          {(lbl71,6);(start,6);(start0,6);(stop,6)}
        Flow Graph:
          [0->{},1->{2,3},2->{},3->{2,3},4->{0,1}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(n^1))