YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. f3(A,B) -> f2(A,B)        True     (1,1)
          1. f2(A,B) -> f2(-1 + A,B)   [A >= 1] (?,1)
          2. f2(A,B) -> f300(-1 + A,C) [0 >= A] (?,1)
        Signature:
          {(f2,2);(f3,2);(f300,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          f3(A,B) -> f2(A,B)         True
          f2(A,B) -> f2(-1 + A,B)    [A >= 1]
          f2(A,B) -> f300(-1 + A,C)  [0 >= A]
        Signature:
          {(f2,2);(f3,2);(f300,2)}
        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:
          f3(A,B) -> f2(A,B)         True
          f2(A,B) -> f2(-1 + A,B)    [A >= 1]
          f2(A,B) -> f300(-1 + A,C)  [0 >= A]
        Signature:
          {(f2,2);(f3,2);(f300,2)}
        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