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

YES