NO
* Step 1: FromIts NO
    + Considered Problem:
        Rules:
          0. f1(A,B,C) -> f1(E,F,B) [A >= 1 && D >= 1 && B = 0] (?,1)
          1. f0(A,B,C) -> f1(A,B,C) True                        (1,1)
        Signature:
          {(f0,3);(f1,3)}
        Flow Graph:
          [0->{0},1->{0}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: CloseWith NO
    + Considered Problem:
        Rules:
          f1(A,B,C) -> f1(E,F,B)  [A >= 1 && D >= 1 && B = 0]
          f0(A,B,C) -> f1(A,B,C)  True
        Signature:
          {(f0,3);(f1,3)}
        Rule Graph:
          [0->{0},1->{0}]
    + Applied Processor:
        CloseWith False
    + Details:
        ()

NO