YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0.  evalfstart(A,B,C,D)    -> evalfentryin(A,B,C,D)               True         (1,1)
          1.  evalfentryin(A,B,C,D)  -> evalfbb5in(B,B,C,D)                 True         (?,1)
          2.  evalfbb5in(A,B,C,D)    -> evalfbbin(A,B,C,D)                  [B >= 2]     (?,1)
          3.  evalfbb5in(A,B,C,D)    -> evalfreturnin(A,B,C,D)              [1 >= B]     (?,1)
          4.  evalfbbin(A,B,C,D)     -> evalfbb2in(A,B,-1 + B,-1 + A + B)   True         (?,1)
          5.  evalfbb2in(A,B,C,D)    -> evalfbb4in(A,B,C,D)                 [C >= 1 + D] (?,1)
          6.  evalfbb2in(A,B,C,D)    -> evalfbb3in(A,B,C,D)                 [D >= C]     (?,1)
          7.  evalfbb3in(A,B,C,D)    -> evalfbb1in(A,B,C,D)                 [0 >= 1 + E] (?,1)
          8.  evalfbb3in(A,B,C,D)    -> evalfbb1in(A,B,C,D)                 [E >= 1]     (?,1)
          9.  evalfbb3in(A,B,C,D)    -> evalfbb4in(A,B,C,D)                 True         (?,1)
          10. evalfbb1in(A,B,C,D)    -> evalfbb2in(A,B,C,-1 + D)            True         (?,1)
          11. evalfbb4in(A,B,C,D)    -> evalfbb5in(1 + -1*C + D,-1 + C,C,D) True         (?,1)
          12. evalfreturnin(A,B,C,D) -> evalfstop(A,B,C,D)                  True         (?,1)
        Signature:
          {(evalfbb1in,4)
          ;(evalfbb2in,4)
          ;(evalfbb3in,4)
          ;(evalfbb4in,4)
          ;(evalfbb5in,4)
          ;(evalfbbin,4)
          ;(evalfentryin,4)
          ;(evalfreturnin,4)
          ;(evalfstart,4)
          ;(evalfstop,4)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4},3->{12},4->{5,6},5->{11},6->{7,8,9},7->{10},8->{10},9->{11},10->{5,6},11->{2,3}
          ,12->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          evalfstart(A,B,C,D)    -> evalfentryin(A,B,C,D)                 True
          evalfentryin(A,B,C,D)  -> evalfbb5in(B,B,C,D)                   True
          evalfbb5in(A,B,C,D)    -> evalfbbin(A,B,C,D)                    [B >= 2]
          evalfbb5in(A,B,C,D)    -> evalfreturnin(A,B,C,D)                [1 >= B]
          evalfbbin(A,B,C,D)     -> evalfbb2in(A,B,-1 + B,-1 + A + B)     True
          evalfbb2in(A,B,C,D)    -> evalfbb4in(A,B,C,D)                   [C >= 1 + D]
          evalfbb2in(A,B,C,D)    -> evalfbb3in(A,B,C,D)                   [D >= C]
          evalfbb3in(A,B,C,D)    -> evalfbb1in(A,B,C,D)                   [0 >= 1 + E]
          evalfbb3in(A,B,C,D)    -> evalfbb1in(A,B,C,D)                   [E >= 1]
          evalfbb3in(A,B,C,D)    -> evalfbb4in(A,B,C,D)                   True
          evalfbb1in(A,B,C,D)    -> evalfbb2in(A,B,C,-1 + D)              True
          evalfbb4in(A,B,C,D)    -> evalfbb5in(1 + -1*C + D,-1 + C,C,D)   True
          evalfreturnin(A,B,C,D) -> evalfstop(A,B,C,D)                    True
        Signature:
          {(evalfbb1in,4)
          ;(evalfbb2in,4)
          ;(evalfbb3in,4)
          ;(evalfbb4in,4)
          ;(evalfbb5in,4)
          ;(evalfbbin,4)
          ;(evalfentryin,4)
          ;(evalfreturnin,4)
          ;(evalfstart,4)
          ;(evalfstop,4)}
        Rule Graph:
          [0->{1},1->{2,3},2->{4},3->{12},4->{5,6},5->{11},6->{7,8,9},7->{10},8->{10},9->{11},10->{5,6},11->{2,3}
          ,12->{}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8,9,10,11,12]
          |
          `- p:[2,11,5,4,10,7,6,8,9] c: [2,4,5,9,11]
             |
             `- p:[6,10,7,8] c: [6,7,8,10]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          evalfstart(A,B,C,D)    -> evalfentryin(A,B,C,D)                 True
          evalfentryin(A,B,C,D)  -> evalfbb5in(B,B,C,D)                   True
          evalfbb5in(A,B,C,D)    -> evalfbbin(A,B,C,D)                    [B >= 2]
          evalfbb5in(A,B,C,D)    -> evalfreturnin(A,B,C,D)                [1 >= B]
          evalfbbin(A,B,C,D)     -> evalfbb2in(A,B,-1 + B,-1 + A + B)     True
          evalfbb2in(A,B,C,D)    -> evalfbb4in(A,B,C,D)                   [C >= 1 + D]
          evalfbb2in(A,B,C,D)    -> evalfbb3in(A,B,C,D)                   [D >= C]
          evalfbb3in(A,B,C,D)    -> evalfbb1in(A,B,C,D)                   [0 >= 1 + E]
          evalfbb3in(A,B,C,D)    -> evalfbb1in(A,B,C,D)                   [E >= 1]
          evalfbb3in(A,B,C,D)    -> evalfbb4in(A,B,C,D)                   True
          evalfbb1in(A,B,C,D)    -> evalfbb2in(A,B,C,-1 + D)              True
          evalfbb4in(A,B,C,D)    -> evalfbb5in(1 + -1*C + D,-1 + C,C,D)   True
          evalfreturnin(A,B,C,D) -> evalfstop(A,B,C,D)                    True
        Signature:
          {(evalfbb1in,4)
          ;(evalfbb2in,4)
          ;(evalfbb3in,4)
          ;(evalfbb4in,4)
          ;(evalfbb5in,4)
          ;(evalfbbin,4)
          ;(evalfentryin,4)
          ;(evalfreturnin,4)
          ;(evalfstart,4)
          ;(evalfstop,4)}
        Rule Graph:
          [0->{1},1->{2,3},2->{4},3->{12},4->{5,6},5->{11},6->{7,8,9},7->{10},8->{10},9->{11},10->{5,6},11->{2,3}
          ,12->{}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8,9,10,11,12]
          |
          `- p:[2,11,5,4,10,7,6,8,9] c: [2,4,5,9,11]
             |
             `- p:[6,10,7,8] c: [6,7,8,10])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES