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

YES