YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0. evalaxstart(A,B,C)    -> evalaxentryin(A,B,C)   True                                                                                                        (1,1)
          1. evalaxentryin(A,B,C)  -> evalaxbbin(0,B,C)      True                                                                                                        (?,1)
          2. evalaxbbin(A,B,C)     -> evalaxbb2in(A,0,C)     [A >= 0]                                                                                                    (?,1)
          3. evalaxbb2in(A,B,C)    -> evalaxbb3in(A,B,C)     [B >= 0 && A + B >= 0 && A >= 0 && 1 + B >= C]                                                              (?,1)
          4. evalaxbb2in(A,B,C)    -> evalaxbb1in(A,B,C)     [B >= 0 && A + B >= 0 && A >= 0 && C >= 2 + B]                                                              (?,1)
          5. evalaxbb3in(A,B,C)    -> evalaxreturnin(A,B,C)  [1 + B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && 2 + A >= C]                                         (?,1)
          6. evalaxbb3in(A,B,C)    -> evalaxbbin(1 + A,B,C)  [1 + B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && 1 + B >= C && C >= 3 + A]                           (?,1)
          7. evalaxbb1in(A,B,C)    -> evalaxbb2in(A,1 + B,C) [-2 + C >= 0 && -2 + B + C >= 0 && -2 + -1*B + C >= 0 && -2 + A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0] (?,1)
          8. evalaxreturnin(A,B,C) -> evalaxstop(A,B,C)      [1 + B + -1*C >= 0 && 2 + A + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0]                                  (?,1)
        Signature:
          {(evalaxbb1in,3)
          ;(evalaxbb2in,3)
          ;(evalaxbb3in,3)
          ;(evalaxbbin,3)
          ;(evalaxentryin,3)
          ;(evalaxreturnin,3)
          ;(evalaxstart,3)
          ;(evalaxstop,3)}
        Flow Graph:
          [0->{1},1->{2},2->{3,4},3->{5,6},4->{7},5->{8},6->{2},7->{3,4},8->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: Looptree YES
    + Considered Problem:
        Rules:
          0. evalaxstart(A,B,C)    -> evalaxentryin(A,B,C)   True                                                                                                        (1,1)
          1. evalaxentryin(A,B,C)  -> evalaxbbin(0,B,C)      True                                                                                                        (1,1)
          2. evalaxbbin(A,B,C)     -> evalaxbb2in(A,0,C)     [A >= 0]                                                                                                    (?,1)
          3. evalaxbb2in(A,B,C)    -> evalaxbb3in(A,B,C)     [B >= 0 && A + B >= 0 && A >= 0 && 1 + B >= C]                                                              (?,1)
          4. evalaxbb2in(A,B,C)    -> evalaxbb1in(A,B,C)     [B >= 0 && A + B >= 0 && A >= 0 && C >= 2 + B]                                                              (?,1)
          5. evalaxbb3in(A,B,C)    -> evalaxreturnin(A,B,C)  [1 + B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && 2 + A >= C]                                         (1,1)
          6. evalaxbb3in(A,B,C)    -> evalaxbbin(1 + A,B,C)  [1 + B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && 1 + B >= C && C >= 3 + A]                           (?,1)
          7. evalaxbb1in(A,B,C)    -> evalaxbb2in(A,1 + B,C) [-2 + C >= 0 && -2 + B + C >= 0 && -2 + -1*B + C >= 0 && -2 + A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0] (?,1)
          8. evalaxreturnin(A,B,C) -> evalaxstop(A,B,C)      [1 + B + -1*C >= 0 && 2 + A + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0]                                  (1,1)
        Signature:
          {(evalaxbb1in,3)
          ;(evalaxbb2in,3)
          ;(evalaxbb3in,3)
          ;(evalaxbbin,3)
          ;(evalaxentryin,3)
          ;(evalaxreturnin,3)
          ;(evalaxstart,3)
          ;(evalaxstop,3)}
        Flow Graph:
          [0->{1},1->{2},2->{3,4},3->{5,6},4->{7},5->{8},6->{2},7->{3,4},8->{}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8]
          |
          `- p:[2,6,3,7,4] c: [6]
             |
             `- p:[4,7] c: [7]

YES