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

YES