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

YES