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) -> l1(1 + A,-1 + B,C,D) [B >= 1]           (?,1)
          2. l1(A,B,C,D) -> l2(A,B,A,D)          [0 >= B]           (?,1)
          3. l2(A,B,C,D) -> l3(A,B,C,C)          [C >= 1]           (?,1)
          4. l3(A,B,C,D) -> l3(A,B,C,-1 + D)     [D >= 1 && C >= 1] (?,1)
          5. l3(A,B,C,D) -> l2(A,B,-1 + C,D)     [0 >= D && C >= 1] (?,1)
        Signature:
          {(l0,4);(l1,4);(l2,4);(l3,4)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{3},3->{4,5},4->{4,5},5->{3}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(3,5)]
* 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) -> l1(1 + A,-1 + B,C,D) [B >= 1]           (?,1)
          2. l1(A,B,C,D) -> l2(A,B,A,D)          [0 >= B]           (?,1)
          3. l2(A,B,C,D) -> l3(A,B,C,C)          [C >= 1]           (?,1)
          4. l3(A,B,C,D) -> l3(A,B,C,-1 + D)     [D >= 1 && C >= 1] (?,1)
          5. l3(A,B,C,D) -> l2(A,B,-1 + C,D)     [0 >= D && C >= 1] (?,1)
        Signature:
          {(l0,4);(l1,4);(l2,4);(l3,4)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{3},3->{4},4->{4,5},5->{3}]
        
    + 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) -> l1(1 + A,-1 + B,C,D)  [B >= 1]
          l1(A,B,C,D) -> l2(A,B,A,D)           [0 >= B]
          l2(A,B,C,D) -> l3(A,B,C,C)           [C >= 1]
          l3(A,B,C,D) -> l3(A,B,C,-1 + D)      [D >= 1 && C >= 1]
          l3(A,B,C,D) -> l2(A,B,-1 + C,D)      [0 >= D && C >= 1]
        Signature:
          {(l0,4);(l1,4);(l2,4);(l3,4)}
        Rule Graph:
          [0->{1,2},1->{1,2},2->{3},3->{4},4->{4,5},5->{3}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          +- p:[1] c: [1]
          |
          `- p:[3,5,4] c: [3,5]
             |
             `- p:[4] c: [4]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          l0(A,B,C,D) -> l1(0,B,C,D)           True
          l1(A,B,C,D) -> l1(1 + A,-1 + B,C,D)  [B >= 1]
          l1(A,B,C,D) -> l2(A,B,A,D)           [0 >= B]
          l2(A,B,C,D) -> l3(A,B,C,C)           [C >= 1]
          l3(A,B,C,D) -> l3(A,B,C,-1 + D)      [D >= 1 && C >= 1]
          l3(A,B,C,D) -> l2(A,B,-1 + C,D)      [0 >= D && C >= 1]
        Signature:
          {(l0,4);(l1,4);(l2,4);(l3,4)}
        Rule Graph:
          [0->{1,2},1->{1,2},2->{3},3->{4},4->{4,5},5->{3}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          +- p:[1] c: [1]
          |
          `- p:[3,5,4] c: [3,5]
             |
             `- p:[4] c: [4])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES