YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. f1(A,B) -> f300(A,C)    [0 >= 1 + A] (?,1)
          1. f1(A,B) -> f1(-1 + A,B) [A >= 0]     (?,1)
          2. f3(A,B) -> f1(A,B)      True         (1,1)
        Signature:
          {(f1,2);(f3,2);(f300,2)}
        Flow Graph:
          [0->{},1->{0,1},2->{0,1}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          f1(A,B) -> f300(A,C)     [0 >= 1 + A]
          f1(A,B) -> f1(-1 + A,B)  [A >= 0]
          f3(A,B) -> f1(A,B)       True
        Signature:
          {(f1,2);(f3,2);(f300,2)}
        Rule Graph:
          [0->{},1->{0,1},2->{0,1}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2]
          |
          `- p:[1] c: [1]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          f1(A,B) -> f300(A,C)     [0 >= 1 + A]
          f1(A,B) -> f1(-1 + A,B)  [A >= 0]
          f3(A,B) -> f1(A,B)       True
        Signature:
          {(f1,2);(f3,2);(f300,2)}
        Rule Graph:
          [0->{},1->{0,1},2->{0,1}]
        ,We construct a looptree:
          P: [0,1,2]
          |
          `- p:[1] c: [1])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES