YES
* Step 1: TrivialSCCs 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)     [-1 + A + -1*B >= 0 && C >= 1 + D]         (?,1)
          5. evalDis1bbin(A,B,C,D)     -> evalDis1bb2in(A,B,C,D)     [-1 + A + -1*B >= 0 && D >= C]             (?,1)
          6. evalDis1bb1in(A,B,C,D)    -> evalDis1bb3in(A,B,C,1 + D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] (?,1)
          7. evalDis1bb2in(A,B,C,D)    -> evalDis1bb3in(A,1 + B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0]      (?,1)
          8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D)      [-1*A + B >= 0]                            (?,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:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: UnsatPaths 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                                       (1,1)
          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]                                   (1,1)
          4. evalDis1bbin(A,B,C,D)     -> evalDis1bb1in(A,B,C,D)     [-1 + A + -1*B >= 0 && C >= 1 + D]         (?,1)
          5. evalDis1bbin(A,B,C,D)     -> evalDis1bb2in(A,B,C,D)     [-1 + A + -1*B >= 0 && D >= C]             (?,1)
          6. evalDis1bb1in(A,B,C,D)    -> evalDis1bb3in(A,B,C,1 + D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] (?,1)
          7. evalDis1bb2in(A,B,C,D)    -> evalDis1bb3in(A,1 + B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0]      (?,1)
          8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D)      [-1*A + B >= 0]                            (1,1)
        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:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(6,3)]
* Step 3: Looptree 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                                       (1,1)
          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]                                   (1,1)
          4. evalDis1bbin(A,B,C,D)     -> evalDis1bb1in(A,B,C,D)     [-1 + A + -1*B >= 0 && C >= 1 + D]         (?,1)
          5. evalDis1bbin(A,B,C,D)     -> evalDis1bb2in(A,B,C,D)     [-1 + A + -1*B >= 0 && D >= C]             (?,1)
          6. evalDis1bb1in(A,B,C,D)    -> evalDis1bb3in(A,B,C,1 + D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] (?,1)
          7. evalDis1bb2in(A,B,C,D)    -> evalDis1bb3in(A,1 + B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0]      (?,1)
          8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D)      [-1*A + B >= 0]                            (1,1)
        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},7->{2,3},8->{}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8]
          |
          `- p:[2,6,4,7,5] c: [7]
             |
             `- p:[2,6,4] c: [6]

YES