NO
* Step 1: FromIts NO
    + Considered Problem:
        Rules:
          0. f3(A,B) -> f0(2,B)      True     (1,1)
          1. f0(A,B) -> f0(1 + A,C)  [C >= 1] (?,1)
          2. f0(A,B) -> f0(-1 + A,C) [0 >= C] (?,1)
        Signature:
          {(f0,2);(f3,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{1,2}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: CloseWith NO
    + Considered Problem:
        Rules:
          f3(A,B) -> f0(2,B)       True
          f0(A,B) -> f0(1 + A,C)   [C >= 1]
          f0(A,B) -> f0(-1 + A,C)  [0 >= C]
        Signature:
          {(f0,2);(f3,2)}
        Rule Graph:
          [0->{1,2},1->{1,2},2->{1,2}]
    + Applied Processor:
        CloseWith False
    + Details:
        ()

NO