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)          [-1 + B >= 0] (?,1)
          2. l2(A,B,C,D) -> l2(A,B,1 + C,C + D)  [-1 + B >= C] (?,1)
          3. l2(A,B,C,D) -> l1(A + D,-1 + B,C,D) [C >= B]      (?,1)
          4. l1(A,B,C,D) -> l3(A,B,C,D)          [0 >= B]      (?,1)
          5. l3(A,B,C,D) -> l3(-1 + A,B,C,D)     [-1 + A >= 0] (?,1)
        Signature:
          {(l0,4);(l1,4);(l2,4);(l3,4)}
        Flow Graph:
          [0->{1,4},1->{2,3},2->{2,3},3->{1,4},4->{5},5->{5}]
        
    + 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)          [-1 + B >= 0] (?,1)
          2. l2(A,B,C,D) -> l2(A,B,1 + C,C + D)  [-1 + B >= C] (?,1)
          3. l2(A,B,C,D) -> l1(A + D,-1 + B,C,D) [C >= B]      (?,1)
          4. l1(A,B,C,D) -> l3(A,B,C,D)          [0 >= B]      (?,1)
          5. l3(A,B,C,D) -> l3(-1 + A,B,C,D)     [-1 + A >= 0] (?,1)
        Signature:
          {(l0,4);(l1,4);(l2,4);(l3,4)}
        Flow Graph:
          [0->{1,4},1->{2},2->{2,3},3->{1,4},4->{5},5->{5}]
        
    + 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)           [-1 + B >= 0]
          l2(A,B,C,D) -> l2(A,B,1 + C,C + D)   [-1 + B >= C]
          l2(A,B,C,D) -> l1(A + D,-1 + B,C,D)  [C >= B]
          l1(A,B,C,D) -> l3(A,B,C,D)           [0 >= B]
          l3(A,B,C,D) -> l3(-1 + A,B,C,D)      [-1 + A >= 0]
        Signature:
          {(l0,4);(l1,4);(l2,4);(l3,4)}
        Rule Graph:
          [0->{1,4},1->{2},2->{2,3},3->{1,4},4->{5},5->{5}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          +- p:[1,3,2] c: [1,3]
          |  |
          |  `- p:[2] c: [2]
          |
          `- p:[5] c: [5]
* 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)           [-1 + B >= 0]
          l2(A,B,C,D) -> l2(A,B,1 + C,C + D)   [-1 + B >= C]
          l2(A,B,C,D) -> l1(A + D,-1 + B,C,D)  [C >= B]
          l1(A,B,C,D) -> l3(A,B,C,D)           [0 >= B]
          l3(A,B,C,D) -> l3(-1 + A,B,C,D)      [-1 + A >= 0]
        Signature:
          {(l0,4);(l1,4);(l2,4);(l3,4)}
        Rule Graph:
          [0->{1,4},1->{2},2->{2,3},3->{1,4},4->{5},5->{5}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          +- p:[1,3,2] c: [1,3]
          |  |
          |  `- p:[2] c: [2]
          |
          `- p:[5] c: [5])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES