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

YES