YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. start(A,B,C,D)  -> stop(A,B,C,D)       [1 >= A && B = C && D = A]                (?,1)
          1. start(A,B,C,D)  -> lbl32(A,B,C,-1 + D) [A >= 2 && B = C && D = A]                (?,1)
          2. lbl32(A,B,C,D)  -> stop(A,B,C,D)       [A >= 2 && D = 1 && B = C]                (?,1)
          3. lbl32(A,B,C,D)  -> lbl32(A,B,C,-1 + D) [D >= 2 && D >= 1 && A >= 1 + D && B = C] (?,1)
          4. start0(A,B,C,D) -> start(A,C,C,A)      True                                      (1,1)
        Signature:
          {(lbl32,4);(start,4);(start0,4);(stop,4)}
        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)  -> stop(A,B,C,D)        [1 >= A && B = C && D = A]
          start(A,B,C,D)  -> lbl32(A,B,C,-1 + D)  [A >= 2 && B = C && D = A]
          lbl32(A,B,C,D)  -> stop(A,B,C,D)        [A >= 2 && D = 1 && B = C]
          lbl32(A,B,C,D)  -> lbl32(A,B,C,-1 + D)  [D >= 2 && D >= 1 && A >= 1 + D && B = C]
          start0(A,B,C,D) -> start(A,C,C,A)       True
        Signature:
          {(lbl32,4);(start,4);(start0,4);(stop,4)}
        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)  -> stop(A,B,C,D)        [1 >= A && B = C && D = A]
          start(A,B,C,D)  -> lbl32(A,B,C,-1 + D)  [A >= 2 && B = C && D = A]
          lbl32(A,B,C,D)  -> stop(A,B,C,D)        [A >= 2 && D = 1 && B = C]
          lbl32(A,B,C,D)  -> lbl32(A,B,C,-1 + D)  [D >= 2 && D >= 1 && A >= 1 + D && B = C]
          start0(A,B,C,D) -> start(A,C,C,A)       True
        Signature:
          {(lbl32,4);(start,4);(start0,4);(stop,4)}
        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