YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. evalndloopstart(A)    -> evalndloopentryin(A)  True                                           (1,1)
          1. evalndloopentryin(A)  -> evalndloopbbin(0)     True                                           (?,1)
          2. evalndloopbbin(A)     -> evalndloopbbin(B)     [A >= 0 && 2 + A >= B && B >= 1 + A && 9 >= B] (?,1)
          3. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && B >= 3 + A]                         (?,1)
          4. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && A >= B]                             (?,1)
          5. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && B >= 10]                            (?,1)
          6. evalndloopreturnin(A) -> evalndloopstop(A)     [A >= 0]                                       (?,1)
        Signature:
          {(evalndloopbbin,1);(evalndloopentryin,1);(evalndloopreturnin,1);(evalndloopstart,1);(evalndloopstop,1)}
        Flow Graph:
          [0->{1},1->{2,3,4,5},2->{2,3,4,5},3->{6},4->{6},5->{6},6->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          evalndloopstart(A)    -> evalndloopentryin(A)   True
          evalndloopentryin(A)  -> evalndloopbbin(0)      True
          evalndloopbbin(A)     -> evalndloopbbin(B)      [A >= 0 && 2 + A >= B && B >= 1 + A && 9 >= B]
          evalndloopbbin(A)     -> evalndloopreturnin(A)  [A >= 0 && B >= 3 + A]
          evalndloopbbin(A)     -> evalndloopreturnin(A)  [A >= 0 && A >= B]
          evalndloopbbin(A)     -> evalndloopreturnin(A)  [A >= 0 && B >= 10]
          evalndloopreturnin(A) -> evalndloopstop(A)      [A >= 0]
        Signature:
          {(evalndloopbbin,1);(evalndloopentryin,1);(evalndloopreturnin,1);(evalndloopstart,1);(evalndloopstop,1)}
        Rule Graph:
          [0->{1},1->{2,3,4,5},2->{2,3,4,5},3->{6},4->{6},5->{6},6->{}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6]
          |
          `- p:[2] c: [2]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          evalndloopstart(A)    -> evalndloopentryin(A)   True
          evalndloopentryin(A)  -> evalndloopbbin(0)      True
          evalndloopbbin(A)     -> evalndloopbbin(B)      [A >= 0 && 2 + A >= B && B >= 1 + A && 9 >= B]
          evalndloopbbin(A)     -> evalndloopreturnin(A)  [A >= 0 && B >= 3 + A]
          evalndloopbbin(A)     -> evalndloopreturnin(A)  [A >= 0 && A >= B]
          evalndloopbbin(A)     -> evalndloopreturnin(A)  [A >= 0 && B >= 10]
          evalndloopreturnin(A) -> evalndloopstop(A)      [A >= 0]
        Signature:
          {(evalndloopbbin,1);(evalndloopentryin,1);(evalndloopreturnin,1);(evalndloopstart,1);(evalndloopstop,1)}
        Rule Graph:
          [0->{1},1->{2,3,4,5},2->{2,3,4,5},3->{6},4->{6},5->{6},6->{}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5,6]
          |
          `- p:[2] c: [2])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES