YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. l0(A,B,C,D) -> l1(0,B,C,D)          True                (1,1)
          1. l1(A,B,C,D) -> l2(A,B,0,0)          [A >= 0 && B >= 1]  (?,1)
          2. l2(A,B,C,D) -> l2(A,B,1 + C,C + D)  [D >= 0             (?,1)
                                                  && C + D >= 0           
                                                  && -1 + B + D >= 0      
                                                  && A + D >= 0           
                                                  && C >= 0               
                                                  && -1 + B + C >= 0      
                                                  && A + C >= 0           
                                                  && -1 + B >= 0          
                                                  && -1 + A + B >= 0      
                                                  && A >= 0               
                                                  && B >= 1 + C]          
          3. l2(A,B,C,D) -> l1(A + D,-1 + B,C,D) [D >= 0             (?,1)
                                                  && C + D >= 0           
                                                  && -1 + B + D >= 0      
                                                  && A + D >= 0           
                                                  && C >= 0               
                                                  && -1 + B + C >= 0      
                                                  && A + C >= 0           
                                                  && -1 + B >= 0          
                                                  && -1 + A + B >= 0      
                                                  && A >= 0               
                                                  && C >= B]              
        Signature:
          {(l0,4);(l1,4);(l2,4)}
        Flow Graph:
          [0->{1},1->{2,3},2->{2,3},3->{1}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(1,3)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0. l0(A,B,C,D) -> l1(0,B,C,D)          True                (1,1)
          1. l1(A,B,C,D) -> l2(A,B,0,0)          [A >= 0 && B >= 1]  (?,1)
          2. l2(A,B,C,D) -> l2(A,B,1 + C,C + D)  [D >= 0             (?,1)
                                                  && C + D >= 0           
                                                  && -1 + B + D >= 0      
                                                  && A + D >= 0           
                                                  && C >= 0               
                                                  && -1 + B + C >= 0      
                                                  && A + C >= 0           
                                                  && -1 + B >= 0          
                                                  && -1 + A + B >= 0      
                                                  && A >= 0               
                                                  && B >= 1 + C]          
          3. l2(A,B,C,D) -> l1(A + D,-1 + B,C,D) [D >= 0             (?,1)
                                                  && C + D >= 0           
                                                  && -1 + B + D >= 0      
                                                  && A + D >= 0           
                                                  && C >= 0               
                                                  && -1 + B + C >= 0      
                                                  && A + C >= 0           
                                                  && -1 + B >= 0          
                                                  && -1 + A + B >= 0      
                                                  && A >= 0               
                                                  && C >= B]              
        Signature:
          {(l0,4);(l1,4);(l2,4)}
        Flow Graph:
          [0->{1},1->{2},2->{2,3},3->{1}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          l0(A,B,C,D) -> l1(0,B,C,D)           True
          l1(A,B,C,D) -> l2(A,B,0,0)           [A >= 0 && B >= 1]
          l2(A,B,C,D) -> l2(A,B,1 + C,C + D)   [D >= 0
                                                && C + D >= 0
                                                && -1 + B + D >= 0
                                                && A + D >= 0
                                                && C >= 0
                                                && -1 + B + C >= 0
                                                && A + C >= 0
                                                && -1 + B >= 0
                                                && -1 + A + B >= 0
                                                && A >= 0
                                                && B >= 1 + C]
          l2(A,B,C,D) -> l1(A + D,-1 + B,C,D)  [D >= 0
                                                && C + D >= 0
                                                && -1 + B + D >= 0
                                                && A + D >= 0
                                                && C >= 0
                                                && -1 + B + C >= 0
                                                && A + C >= 0
                                                && -1 + B >= 0
                                                && -1 + A + B >= 0
                                                && A >= 0
                                                && C >= B]
        Signature:
          {(l0,4);(l1,4);(l2,4)}
        Rule Graph:
          [0->{1},1->{2},2->{2,3},3->{1}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[1,3,2] c: [1,3]
             |
             `- p:[2] c: [2]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          l0(A,B,C,D) -> l1(0,B,C,D)           True
          l1(A,B,C,D) -> l2(A,B,0,0)           [A >= 0 && B >= 1]
          l2(A,B,C,D) -> l2(A,B,1 + C,C + D)   [D >= 0
                                                && C + D >= 0
                                                && -1 + B + D >= 0
                                                && A + D >= 0
                                                && C >= 0
                                                && -1 + B + C >= 0
                                                && A + C >= 0
                                                && -1 + B >= 0
                                                && -1 + A + B >= 0
                                                && A >= 0
                                                && B >= 1 + C]
          l2(A,B,C,D) -> l1(A + D,-1 + B,C,D)  [D >= 0
                                                && C + D >= 0
                                                && -1 + B + D >= 0
                                                && A + D >= 0
                                                && C >= 0
                                                && -1 + B + C >= 0
                                                && A + C >= 0
                                                && -1 + B >= 0
                                                && -1 + A + B >= 0
                                                && A >= 0
                                                && C >= B]
        Signature:
          {(l0,4);(l1,4);(l2,4)}
        Rule Graph:
          [0->{1},1->{2},2->{2,3},3->{1}]
        ,We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[1,3,2] c: [1,3]
             |
             `- p:[2] c: [2])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES