NO
* Step 1: TrivialSCCs NO
    + Considered Problem:
        Rules:
          0.  evalEx3start(A,B,C)    -> evalEx3entryin(A,B,C)    True                                                                           (1,1)
          1.  evalEx3entryin(A,B,C)  -> evalEx3bb4in(A,B,C)      True                                                                           (?,1)
          2.  evalEx3bb4in(A,B,C)    -> evalEx3bbin(A,B,C)       [A >= 1]                                                                       (?,1)
          3.  evalEx3bb4in(A,B,C)    -> evalEx3returnin(A,B,C)   [0 >= A]                                                                       (?,1)
          4.  evalEx3bbin(A,B,C)     -> evalEx3bb2in(A,D,A)      [-1 + A >= 0]                                                                  (?,1)
          5.  evalEx3bb2in(A,B,C)    -> evalEx3bb4in(C,B,C)      [A + -1*C >= 0 && -1 + A >= 0 && 0 >= C]                                       (?,1)
          6.  evalEx3bb2in(A,B,C)    -> evalEx3bb3in(A,B,C)      [A + -1*C >= 0 && -1 + A >= 0 && C >= 1]                                       (?,1)
          7.  evalEx3bb3in(A,B,C)    -> evalEx3bb1in(A,B,C)      [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0]               (?,1)
          8.  evalEx3bb3in(A,B,C)    -> evalEx3bb4in(C,B,C)      [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0 && B >= 1 + D] (?,1)
          9.  evalEx3bb3in(A,B,C)    -> evalEx3bb4in(C,B,C)      [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0 && D >= 1 + B] (?,1)
          10. evalEx3bb1in(A,B,C)    -> evalEx3bb2in(A,B,-1 + C) [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0]               (?,1)
          11. evalEx3returnin(A,B,C) -> evalEx3stop(A,B,C)       [-1*A >= 0]                                                                    (?,1)
        Signature:
          {(evalEx3bb1in,3)
          ;(evalEx3bb2in,3)
          ;(evalEx3bb3in,3)
          ;(evalEx3bb4in,3)
          ;(evalEx3bbin,3)
          ;(evalEx3entryin,3)
          ;(evalEx3returnin,3)
          ;(evalEx3start,3)
          ;(evalEx3stop,3)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4},3->{11},4->{5,6},5->{2,3},6->{7,8,9},7->{10},8->{2,3},9->{2,3},10->{5,6},11->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: UnsatPaths NO
    + Considered Problem:
        Rules:
          0.  evalEx3start(A,B,C)    -> evalEx3entryin(A,B,C)    True                                                                           (1,1)
          1.  evalEx3entryin(A,B,C)  -> evalEx3bb4in(A,B,C)      True                                                                           (1,1)
          2.  evalEx3bb4in(A,B,C)    -> evalEx3bbin(A,B,C)       [A >= 1]                                                                       (?,1)
          3.  evalEx3bb4in(A,B,C)    -> evalEx3returnin(A,B,C)   [0 >= A]                                                                       (1,1)
          4.  evalEx3bbin(A,B,C)     -> evalEx3bb2in(A,D,A)      [-1 + A >= 0]                                                                  (?,1)
          5.  evalEx3bb2in(A,B,C)    -> evalEx3bb4in(C,B,C)      [A + -1*C >= 0 && -1 + A >= 0 && 0 >= C]                                       (?,1)
          6.  evalEx3bb2in(A,B,C)    -> evalEx3bb3in(A,B,C)      [A + -1*C >= 0 && -1 + A >= 0 && C >= 1]                                       (?,1)
          7.  evalEx3bb3in(A,B,C)    -> evalEx3bb1in(A,B,C)      [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0]               (?,1)
          8.  evalEx3bb3in(A,B,C)    -> evalEx3bb4in(C,B,C)      [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0 && B >= 1 + D] (?,1)
          9.  evalEx3bb3in(A,B,C)    -> evalEx3bb4in(C,B,C)      [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0 && D >= 1 + B] (?,1)
          10. evalEx3bb1in(A,B,C)    -> evalEx3bb2in(A,B,-1 + C) [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0]               (?,1)
          11. evalEx3returnin(A,B,C) -> evalEx3stop(A,B,C)       [-1*A >= 0]                                                                    (1,1)
        Signature:
          {(evalEx3bb1in,3)
          ;(evalEx3bb2in,3)
          ;(evalEx3bb3in,3)
          ;(evalEx3bb4in,3)
          ;(evalEx3bbin,3)
          ;(evalEx3entryin,3)
          ;(evalEx3returnin,3)
          ;(evalEx3start,3)
          ;(evalEx3stop,3)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4},3->{11},4->{5,6},5->{2,3},6->{7,8,9},7->{10},8->{2,3},9->{2,3},10->{5,6},11->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(4,5),(5,2),(8,3),(9,3)]
* Step 3: Looptree NO
    + Considered Problem:
        Rules:
          0.  evalEx3start(A,B,C)    -> evalEx3entryin(A,B,C)    True                                                                           (1,1)
          1.  evalEx3entryin(A,B,C)  -> evalEx3bb4in(A,B,C)      True                                                                           (1,1)
          2.  evalEx3bb4in(A,B,C)    -> evalEx3bbin(A,B,C)       [A >= 1]                                                                       (?,1)
          3.  evalEx3bb4in(A,B,C)    -> evalEx3returnin(A,B,C)   [0 >= A]                                                                       (1,1)
          4.  evalEx3bbin(A,B,C)     -> evalEx3bb2in(A,D,A)      [-1 + A >= 0]                                                                  (?,1)
          5.  evalEx3bb2in(A,B,C)    -> evalEx3bb4in(C,B,C)      [A + -1*C >= 0 && -1 + A >= 0 && 0 >= C]                                       (?,1)
          6.  evalEx3bb2in(A,B,C)    -> evalEx3bb3in(A,B,C)      [A + -1*C >= 0 && -1 + A >= 0 && C >= 1]                                       (?,1)
          7.  evalEx3bb3in(A,B,C)    -> evalEx3bb1in(A,B,C)      [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0]               (?,1)
          8.  evalEx3bb3in(A,B,C)    -> evalEx3bb4in(C,B,C)      [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0 && B >= 1 + D] (?,1)
          9.  evalEx3bb3in(A,B,C)    -> evalEx3bb4in(C,B,C)      [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0 && D >= 1 + B] (?,1)
          10. evalEx3bb1in(A,B,C)    -> evalEx3bb2in(A,B,-1 + C) [A + -1*C >= 0 && -1 + C >= 0 && -2 + A + C >= 0 && -1 + A >= 0]               (?,1)
          11. evalEx3returnin(A,B,C) -> evalEx3stop(A,B,C)       [-1*A >= 0]                                                                    (1,1)
        Signature:
          {(evalEx3bb1in,3)
          ;(evalEx3bb2in,3)
          ;(evalEx3bb3in,3)
          ;(evalEx3bb4in,3)
          ;(evalEx3bbin,3)
          ;(evalEx3entryin,3)
          ;(evalEx3returnin,3)
          ;(evalEx3start,3)
          ;(evalEx3stop,3)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4},3->{11},4->{6},5->{3},6->{7,8,9},7->{10},8->{2},9->{2},10->{5,6},11->{}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8,9,10,11]
          |
          `- p:[2,8,6,4,10,7,9] c: [10]
             |
             `- p:[2,8,6,4,9] c: []

NO