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

YES