YES
* Step 1: FromIts YES
    + 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:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          start(A,B,C,D,E,F)  -> stop(A,B,C,D,E,F)                 [0 >= A && B = A && C = D && E = F]
          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]
          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]
          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]
          start0(A,B,C,D,E,F) -> start(A,A,D,D,F,F)                True
        Signature:
          {(lbl71,6);(start,6);(start0,6);(stop,6)}
        Rule Graph:
          [0->{},1->{2,3},2->{},3->{2,3},4->{0,1}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[3] c: [3]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          start(A,B,C,D,E,F)  -> stop(A,B,C,D,E,F)                 [0 >= A && B = A && C = D && E = F]
          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]
          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]
          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]
          start0(A,B,C,D,E,F) -> start(A,A,D,D,F,F)                True
        Signature:
          {(lbl71,6);(start,6);(start0,6);(stop,6)}
        Rule Graph:
          [0->{},1->{2,3},2->{},3->{2,3},4->{0,1}]
        ,We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[3] c: [3])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES