YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. evalDis1start(A,B,C,D)    -> evalDis1entryin(A,B,C,D)   True         (1,1)
          1. evalDis1entryin(A,B,C,D)  -> evalDis1bb3in(B,A,D,C)     True         (?,1)
          2. evalDis1bb3in(A,B,C,D)    -> evalDis1bbin(A,B,C,D)      [A >= 1 + B] (?,1)
          3. evalDis1bb3in(A,B,C,D)    -> evalDis1returnin(A,B,C,D)  [B >= A]     (?,1)
          4. evalDis1bbin(A,B,C,D)     -> evalDis1bb1in(A,B,C,D)     [C >= 1 + D] (?,1)
          5. evalDis1bbin(A,B,C,D)     -> evalDis1bb2in(A,B,C,D)     [D >= C]     (?,1)
          6. evalDis1bb1in(A,B,C,D)    -> evalDis1bb3in(A,B,C,1 + D) True         (?,1)
          7. evalDis1bb2in(A,B,C,D)    -> evalDis1bb3in(A,1 + B,C,D) True         (?,1)
          8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D)      True         (?,1)
        Signature:
          {(evalDis1bb1in,4)
          ;(evalDis1bb2in,4)
          ;(evalDis1bb3in,4)
          ;(evalDis1bbin,4)
          ;(evalDis1entryin,4)
          ;(evalDis1returnin,4)
          ;(evalDis1start,4)
          ;(evalDis1stop,4)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2,3},7->{2,3},8->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          evalDis1start(A,B,C,D)    -> evalDis1entryin(A,B,C,D)    True
          evalDis1entryin(A,B,C,D)  -> evalDis1bb3in(B,A,D,C)      True
          evalDis1bb3in(A,B,C,D)    -> evalDis1bbin(A,B,C,D)       [A >= 1 + B]
          evalDis1bb3in(A,B,C,D)    -> evalDis1returnin(A,B,C,D)   [B >= A]
          evalDis1bbin(A,B,C,D)     -> evalDis1bb1in(A,B,C,D)      [C >= 1 + D]
          evalDis1bbin(A,B,C,D)     -> evalDis1bb2in(A,B,C,D)      [D >= C]
          evalDis1bb1in(A,B,C,D)    -> evalDis1bb3in(A,B,C,1 + D)  True
          evalDis1bb2in(A,B,C,D)    -> evalDis1bb3in(A,1 + B,C,D)  True
          evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D)       True
        Signature:
          {(evalDis1bb1in,4)
          ;(evalDis1bb2in,4)
          ;(evalDis1bb3in,4)
          ;(evalDis1bbin,4)
          ;(evalDis1entryin,4)
          ;(evalDis1returnin,4)
          ;(evalDis1start,4)
          ;(evalDis1stop,4)}
        Rule Graph:
          [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2,3},7->{2,3},8->{}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8]
          |
          `- p:[2,6,4,7,5] c: [4,6]
             |
             `- p:[2,7,5] c: [2,5,7]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          evalDis1start(A,B,C,D)    -> evalDis1entryin(A,B,C,D)    True
          evalDis1entryin(A,B,C,D)  -> evalDis1bb3in(B,A,D,C)      True
          evalDis1bb3in(A,B,C,D)    -> evalDis1bbin(A,B,C,D)       [A >= 1 + B]
          evalDis1bb3in(A,B,C,D)    -> evalDis1returnin(A,B,C,D)   [B >= A]
          evalDis1bbin(A,B,C,D)     -> evalDis1bb1in(A,B,C,D)      [C >= 1 + D]
          evalDis1bbin(A,B,C,D)     -> evalDis1bb2in(A,B,C,D)      [D >= C]
          evalDis1bb1in(A,B,C,D)    -> evalDis1bb3in(A,B,C,1 + D)  True
          evalDis1bb2in(A,B,C,D)    -> evalDis1bb3in(A,1 + B,C,D)  True
          evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D)       True
        Signature:
          {(evalDis1bb1in,4)
          ;(evalDis1bb2in,4)
          ;(evalDis1bb3in,4)
          ;(evalDis1bbin,4)
          ;(evalDis1entryin,4)
          ;(evalDis1returnin,4)
          ;(evalDis1start,4)
          ;(evalDis1stop,4)}
        Rule Graph:
          [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2,3},7->{2,3},8->{}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8]
          |
          `- p:[2,6,4,7,5] c: [4,6]
             |
             `- p:[2,7,5] c: [2,5,7])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES