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

YES