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