YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0.  evalNestedSinglestart(A,B,C)    -> evalNestedSingleentryin(A,B,C)   True         (1,1)
          1.  evalNestedSingleentryin(A,B,C)  -> evalNestedSinglebb5in(0,B,C)     True         (?,1)
          2.  evalNestedSinglebb5in(A,B,C)    -> evalNestedSinglebb2in(A,B,A)     [B >= 1 + A] (?,1)
          3.  evalNestedSinglebb5in(A,B,C)    -> evalNestedSinglereturnin(A,B,C)  [A >= B]     (?,1)
          4.  evalNestedSinglebb2in(A,B,C)    -> evalNestedSinglebb4in(A,B,C)     [C >= B]     (?,1)
          5.  evalNestedSinglebb2in(A,B,C)    -> evalNestedSinglebb3in(A,B,C)     [B >= 1 + C] (?,1)
          6.  evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb1in(A,B,C)     [0 >= 1 + D] (?,1)
          7.  evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb1in(A,B,C)     [D >= 1]     (?,1)
          8.  evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb4in(A,B,C)     True         (?,1)
          9.  evalNestedSinglebb1in(A,B,C)    -> evalNestedSinglebb2in(A,B,1 + C) True         (?,1)
          10. evalNestedSinglebb4in(A,B,C)    -> evalNestedSinglebb5in(1 + C,B,C) True         (?,1)
          11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C)      True         (?,1)
        Signature:
          {(evalNestedSinglebb1in,3)
          ;(evalNestedSinglebb2in,3)
          ;(evalNestedSinglebb3in,3)
          ;(evalNestedSinglebb4in,3)
          ;(evalNestedSinglebb5in,3)
          ;(evalNestedSingleentryin,3)
          ;(evalNestedSinglereturnin,3)
          ;(evalNestedSinglestart,3)
          ;(evalNestedSinglestop,3)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4,5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(2,4)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0.  evalNestedSinglestart(A,B,C)    -> evalNestedSingleentryin(A,B,C)   True         (1,1)
          1.  evalNestedSingleentryin(A,B,C)  -> evalNestedSinglebb5in(0,B,C)     True         (?,1)
          2.  evalNestedSinglebb5in(A,B,C)    -> evalNestedSinglebb2in(A,B,A)     [B >= 1 + A] (?,1)
          3.  evalNestedSinglebb5in(A,B,C)    -> evalNestedSinglereturnin(A,B,C)  [A >= B]     (?,1)
          4.  evalNestedSinglebb2in(A,B,C)    -> evalNestedSinglebb4in(A,B,C)     [C >= B]     (?,1)
          5.  evalNestedSinglebb2in(A,B,C)    -> evalNestedSinglebb3in(A,B,C)     [B >= 1 + C] (?,1)
          6.  evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb1in(A,B,C)     [0 >= 1 + D] (?,1)
          7.  evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb1in(A,B,C)     [D >= 1]     (?,1)
          8.  evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb4in(A,B,C)     True         (?,1)
          9.  evalNestedSinglebb1in(A,B,C)    -> evalNestedSinglebb2in(A,B,1 + C) True         (?,1)
          10. evalNestedSinglebb4in(A,B,C)    -> evalNestedSinglebb5in(1 + C,B,C) True         (?,1)
          11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C)      True         (?,1)
        Signature:
          {(evalNestedSinglebb1in,3)
          ;(evalNestedSinglebb2in,3)
          ;(evalNestedSinglebb3in,3)
          ;(evalNestedSinglebb4in,3)
          ;(evalNestedSinglebb5in,3)
          ;(evalNestedSingleentryin,3)
          ;(evalNestedSinglereturnin,3)
          ;(evalNestedSinglestart,3)
          ;(evalNestedSinglestop,3)}
        Flow Graph:
          [0->{1},1->{2,3},2->{5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          evalNestedSinglestart(A,B,C)    -> evalNestedSingleentryin(A,B,C)    True
          evalNestedSingleentryin(A,B,C)  -> evalNestedSinglebb5in(0,B,C)      True
          evalNestedSinglebb5in(A,B,C)    -> evalNestedSinglebb2in(A,B,A)      [B >= 1 + A]
          evalNestedSinglebb5in(A,B,C)    -> evalNestedSinglereturnin(A,B,C)   [A >= B]
          evalNestedSinglebb2in(A,B,C)    -> evalNestedSinglebb4in(A,B,C)      [C >= B]
          evalNestedSinglebb2in(A,B,C)    -> evalNestedSinglebb3in(A,B,C)      [B >= 1 + C]
          evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb1in(A,B,C)      [0 >= 1 + D]
          evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb1in(A,B,C)      [D >= 1]
          evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb4in(A,B,C)      True
          evalNestedSinglebb1in(A,B,C)    -> evalNestedSinglebb2in(A,B,1 + C)  True
          evalNestedSinglebb4in(A,B,C)    -> evalNestedSinglebb5in(1 + C,B,C)  True
          evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C)       True
        Signature:
          {(evalNestedSinglebb1in,3)
          ;(evalNestedSinglebb2in,3)
          ;(evalNestedSinglebb3in,3)
          ;(evalNestedSinglebb4in,3)
          ;(evalNestedSinglebb5in,3)
          ;(evalNestedSingleentryin,3)
          ;(evalNestedSinglereturnin,3)
          ;(evalNestedSinglestart,3)
          ;(evalNestedSinglestop,3)}
        Rule Graph:
          [0->{1},1->{2,3},2->{5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8,9,10,11]
          |
          `- p:[2,10,4,9,6,5,7,8] c: [2,4,5,6,7,8,9,10]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          evalNestedSinglestart(A,B,C)    -> evalNestedSingleentryin(A,B,C)    True
          evalNestedSingleentryin(A,B,C)  -> evalNestedSinglebb5in(0,B,C)      True
          evalNestedSinglebb5in(A,B,C)    -> evalNestedSinglebb2in(A,B,A)      [B >= 1 + A]
          evalNestedSinglebb5in(A,B,C)    -> evalNestedSinglereturnin(A,B,C)   [A >= B]
          evalNestedSinglebb2in(A,B,C)    -> evalNestedSinglebb4in(A,B,C)      [C >= B]
          evalNestedSinglebb2in(A,B,C)    -> evalNestedSinglebb3in(A,B,C)      [B >= 1 + C]
          evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb1in(A,B,C)      [0 >= 1 + D]
          evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb1in(A,B,C)      [D >= 1]
          evalNestedSinglebb3in(A,B,C)    -> evalNestedSinglebb4in(A,B,C)      True
          evalNestedSinglebb1in(A,B,C)    -> evalNestedSinglebb2in(A,B,1 + C)  True
          evalNestedSinglebb4in(A,B,C)    -> evalNestedSinglebb5in(1 + C,B,C)  True
          evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C)       True
        Signature:
          {(evalNestedSinglebb1in,3)
          ;(evalNestedSinglebb2in,3)
          ;(evalNestedSinglebb3in,3)
          ;(evalNestedSinglebb4in,3)
          ;(evalNestedSinglebb5in,3)
          ;(evalNestedSingleentryin,3)
          ;(evalNestedSinglereturnin,3)
          ;(evalNestedSinglestart,3)
          ;(evalNestedSinglestop,3)}
        Rule Graph:
          [0->{1},1->{2,3},2->{5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8,9,10,11]
          |
          `- p:[2,10,4,9,6,5,7,8] c: [2,4,5,6,7,8,9,10])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES