YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. evalndecrstart(A)    -> evalndecrentryin(A)    True     (1,1)
          1. evalndecrentryin(A)  -> evalndecrbb1in(-1 + A) True     (?,1)
          2. evalndecrbb1in(A)    -> evalndecrbbin(A)       [A >= 2] (?,1)
          3. evalndecrbb1in(A)    -> evalndecrreturnin(A)   [1 >= A] (?,1)
          4. evalndecrbbin(A)     -> evalndecrbb1in(-1 + A) True     (?,1)
          5. evalndecrreturnin(A) -> evalndecrstop(A)       True     (?,1)
        Signature:
          {(evalndecrbb1in,1)
          ;(evalndecrbbin,1)
          ;(evalndecrentryin,1)
          ;(evalndecrreturnin,1)
          ;(evalndecrstart,1)
          ;(evalndecrstop,1)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          evalndecrstart(A)    -> evalndecrentryin(A)     True
          evalndecrentryin(A)  -> evalndecrbb1in(-1 + A)  True
          evalndecrbb1in(A)    -> evalndecrbbin(A)        [A >= 2]
          evalndecrbb1in(A)    -> evalndecrreturnin(A)    [1 >= A]
          evalndecrbbin(A)     -> evalndecrbb1in(-1 + A)  True
          evalndecrreturnin(A) -> evalndecrstop(A)        True
        Signature:
          {(evalndecrbb1in,1)
          ;(evalndecrbbin,1)
          ;(evalndecrentryin,1)
          ;(evalndecrreturnin,1)
          ;(evalndecrstart,1)
          ;(evalndecrstop,1)}
        Rule Graph:
          [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          `- p:[2,4] c: [2,4]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          evalndecrstart(A)    -> evalndecrentryin(A)     True
          evalndecrentryin(A)  -> evalndecrbb1in(-1 + A)  True
          evalndecrbb1in(A)    -> evalndecrbbin(A)        [A >= 2]
          evalndecrbb1in(A)    -> evalndecrreturnin(A)    [1 >= A]
          evalndecrbbin(A)     -> evalndecrbb1in(-1 + A)  True
          evalndecrreturnin(A) -> evalndecrstop(A)        True
        Signature:
          {(evalndecrbb1in,1)
          ;(evalndecrbbin,1)
          ;(evalndecrentryin,1)
          ;(evalndecrreturnin,1)
          ;(evalndecrstart,1)
          ;(evalndecrstop,1)}
        Rule Graph:
          [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          `- p:[2,4] c: [2,4])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES