YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. f2(A,B)   -> f2(-1 + A,B) [A >= 2] (?,1)
          1. f2(A,B)   -> f1(-1 + A,C) [1 >= A] (?,1)
          2. f300(A,B) -> f2(A,B)      True     (1,1)
        Signature:
          {(f1,2);(f2,2);(f300,2)}
        Flow Graph:
          [0->{0,1},1->{},2->{0,1}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          f2(A,B)   -> f2(-1 + A,B)  [A >= 2]
          f2(A,B)   -> f1(-1 + A,C)  [1 >= A]
          f300(A,B) -> f2(A,B)       True
        Signature:
          {(f1,2);(f2,2);(f300,2)}
        Rule Graph:
          [0->{0,1},1->{},2->{0,1}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2]
          |
          `- p:[0] c: [0]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          f2(A,B)   -> f2(-1 + A,B)  [A >= 2]
          f2(A,B)   -> f1(-1 + A,C)  [1 >= A]
          f300(A,B) -> f2(A,B)       True
        Signature:
          {(f1,2);(f2,2);(f300,2)}
        Rule Graph:
          [0->{0,1},1->{},2->{0,1}]
        ,We construct a looptree:
          P: [0,1,2]
          |
          `- p:[0] c: [0])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES