YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. f0(A,B) -> f1(A,B)        [A >= 1]                          (1,1)
          1. f1(A,B) -> f1(A,-1*A + B) [-1 + A >= 0 && A >= 1 && B >= 0] (?,1)
        Signature:
          {(f0,2);(f1,2)}
        Flow Graph:
          [0->{1},1->{1}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          f0(A,B) -> f1(A,B)         [A >= 1]
          f1(A,B) -> f1(A,-1*A + B)  [-1 + A >= 0 && A >= 1 && B >= 0]
        Signature:
          {(f0,2);(f1,2)}
        Rule Graph:
          [0->{1},1->{1}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1]
          |
          `- p:[1] c: [1]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          f0(A,B) -> f1(A,B)         [A >= 1]
          f1(A,B) -> f1(A,-1*A + B)  [-1 + A >= 0 && A >= 1 && B >= 0]
        Signature:
          {(f0,2);(f1,2)}
        Rule Graph:
          [0->{1},1->{1}]
        ,We construct a looptree:
          P: [0,1]
          |
          `- p:[1] c: [1])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES