YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. f(A,B,C)  -> g(A,1,0)       True          (1,1)
          1. g(A,B,C)  -> g1(-1 + A,B,B) [-1 + A >= 0] (?,1)
          2. g1(A,B,C) -> g(A,B + C,C)   True          (?,1)
          3. g(A,B,C)  -> h(A,B,C)       [0 >= A]      (?,1)
          4. h(A,B,C)  -> h(A,-1 + B,C)  [-1 + B >= 0] (?,1)
        Signature:
          {(f,3);(g,3);(g1,3);(h,3)}
        Flow Graph:
          [0->{1,3},1->{2},2->{1,3},3->{4},4->{4}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          f(A,B,C)  -> g(A,1,0)        True
          g(A,B,C)  -> g1(-1 + A,B,B)  [-1 + A >= 0]
          g1(A,B,C) -> g(A,B + C,C)    True
          g(A,B,C)  -> h(A,B,C)        [0 >= A]
          h(A,B,C)  -> h(A,-1 + B,C)   [-1 + B >= 0]
        Signature:
          {(f,3);(g,3);(g1,3);(h,3)}
        Rule Graph:
          [0->{1,3},1->{2},2->{1,3},3->{4},4->{4}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4]
          |
          +- p:[1,2] c: [1,2]
          |
          `- p:[4] c: [4]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          f(A,B,C)  -> g(A,1,0)        True
          g(A,B,C)  -> g1(-1 + A,B,B)  [-1 + A >= 0]
          g1(A,B,C) -> g(A,B + C,C)    True
          g(A,B,C)  -> h(A,B,C)        [0 >= A]
          h(A,B,C)  -> h(A,-1 + B,C)   [-1 + B >= 0]
        Signature:
          {(f,3);(g,3);(g1,3);(h,3)}
        Rule Graph:
          [0->{1,3},1->{2},2->{1,3},3->{4},4->{4}]
        ,We construct a looptree:
          P: [0,1,2,3,4]
          |
          +- p:[1,2] c: [1,2]
          |
          `- p:[4] c: [4])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES