YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. start(A) -> a(A) [A >= 1]                  (1,1)
          1. start(A) -> a(A) [A >= 2]                  (1,1)
          2. start(A) -> a(A) [A >= 4]                  (1,1)
          3. a(A)     -> a(B) [2*B >= 2 && A = 2*B]     (?,1)
          4. a(A)     -> a(B) [2*B >= 1 && A = 1 + 2*B] (?,1)
        Signature:
          {(a,1);(start,1)}
        Flow Graph:
          [0->{3,4},1->{3,4},2->{3,4},3->{3,4},4->{3,4}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          start(A) -> a(A)  [A >= 1]
          start(A) -> a(A)  [A >= 2]
          start(A) -> a(A)  [A >= 4]
          a(A)     -> a(B)  [2*B >= 2 && A = 2*B]
          a(A)     -> a(B)  [2*B >= 1 && A = 1 + 2*B]
        Signature:
          {(a,1);(start,1)}
        Rule Graph:
          [0->{3,4},1->{3,4},2->{3,4},3->{3,4},4->{3,4}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[3,4] c: [4]
             |
             `- p:[3] c: [3]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          start(A) -> a(A)  [A >= 1]
          start(A) -> a(A)  [A >= 2]
          start(A) -> a(A)  [A >= 4]
          a(A)     -> a(B)  [2*B >= 2 && A = 2*B]
          a(A)     -> a(B)  [2*B >= 1 && A = 1 + 2*B]
        Signature:
          {(a,1);(start,1)}
        Rule Graph:
          [0->{3,4},1->{3,4},2->{3,4},3->{3,4},4->{3,4}]
        ,We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[3,4] c: [4]
             |
             `- p:[3] c: [3])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES