YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0.  eval_nd_loop_start(v_0,v_x_0)  -> eval_nd_loop_bb0_in(v_0,v_x_0) True                                                                   (1,1)
          1.  eval_nd_loop_bb0_in(v_0,v_x_0) -> eval_nd_loop_0(v_0,v_x_0)      True                                                                   (?,1)
          2.  eval_nd_loop_0(v_0,v_x_0)      -> eval_nd_loop_1(v_0,v_x_0)      True                                                                   (?,1)
          3.  eval_nd_loop_1(v_0,v_x_0)      -> eval_nd_loop_2(v_0,v_x_0)      True                                                                   (?,1)
          4.  eval_nd_loop_2(v_0,v_x_0)      -> eval_nd_loop_3(v_0,v_x_0)      True                                                                   (?,1)
          5.  eval_nd_loop_3(v_0,v_x_0)      -> eval_nd_loop_bb1_in(v_0,0)     True                                                                   (?,1)
          6.  eval_nd_loop_bb1_in(v_0,v_x_0) -> eval_nd_loop_4(v_0,v_x_0)      [v_x_0 >= 0]                                                           (?,1)
          7.  eval_nd_loop_4(v_0,v_x_0)      -> eval_nd_loop_5(nondef_0,v_x_0) [v_x_0 >= 0]                                                           (?,1)
          8.  eval_nd_loop_5(v_0,v_x_0)      -> eval_nd_loop_bb1_in(v_0,v_0)   [v_x_0 >= 0 && 2 >= v_0 + -1*v_x_0 && v_0 + -1*v_x_0 >= 1 && 9 >= v_0] (?,1)
          9.  eval_nd_loop_5(v_0,v_x_0)      -> eval_nd_loop_bb2_in(v_0,v_x_0) [v_x_0 >= 0 && -1 + v_0 + -1*v_x_0 >= 2]                               (?,1)
          10. eval_nd_loop_5(v_0,v_x_0)      -> eval_nd_loop_bb2_in(v_0,v_x_0) [v_x_0 >= 0 && 0 >= v_0 + -1*v_x_0]                                    (?,1)
          11. eval_nd_loop_5(v_0,v_x_0)      -> eval_nd_loop_bb2_in(v_0,v_x_0) [v_x_0 >= 0 && v_0 >= 10]                                              (?,1)
          12. eval_nd_loop_bb2_in(v_0,v_x_0) -> eval_nd_loop_stop(v_0,v_x_0)   [v_x_0 >= 0]                                                           (?,1)
        Signature:
          {(eval_nd_loop_0,2)
          ;(eval_nd_loop_1,2)
          ;(eval_nd_loop_2,2)
          ;(eval_nd_loop_3,2)
          ;(eval_nd_loop_4,2)
          ;(eval_nd_loop_5,2)
          ;(eval_nd_loop_bb0_in,2)
          ;(eval_nd_loop_bb1_in,2)
          ;(eval_nd_loop_bb2_in,2)
          ;(eval_nd_loop_start,2)
          ;(eval_nd_loop_stop,2)}
        Flow Graph:
          [0->{1},1->{2},2->{3},3->{4},4->{5},5->{6},6->{7},7->{8,9,10,11},8->{6},9->{12},10->{12},11->{12},12->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: Looptree YES
    + Considered Problem:
        Rules:
          0.  eval_nd_loop_start(v_0,v_x_0)  -> eval_nd_loop_bb0_in(v_0,v_x_0) True                                                                   (1,1)
          1.  eval_nd_loop_bb0_in(v_0,v_x_0) -> eval_nd_loop_0(v_0,v_x_0)      True                                                                   (1,1)
          2.  eval_nd_loop_0(v_0,v_x_0)      -> eval_nd_loop_1(v_0,v_x_0)      True                                                                   (1,1)
          3.  eval_nd_loop_1(v_0,v_x_0)      -> eval_nd_loop_2(v_0,v_x_0)      True                                                                   (1,1)
          4.  eval_nd_loop_2(v_0,v_x_0)      -> eval_nd_loop_3(v_0,v_x_0)      True                                                                   (1,1)
          5.  eval_nd_loop_3(v_0,v_x_0)      -> eval_nd_loop_bb1_in(v_0,0)     True                                                                   (1,1)
          6.  eval_nd_loop_bb1_in(v_0,v_x_0) -> eval_nd_loop_4(v_0,v_x_0)      [v_x_0 >= 0]                                                           (?,1)
          7.  eval_nd_loop_4(v_0,v_x_0)      -> eval_nd_loop_5(nondef_0,v_x_0) [v_x_0 >= 0]                                                           (?,1)
          8.  eval_nd_loop_5(v_0,v_x_0)      -> eval_nd_loop_bb1_in(v_0,v_0)   [v_x_0 >= 0 && 2 >= v_0 + -1*v_x_0 && v_0 + -1*v_x_0 >= 1 && 9 >= v_0] (?,1)
          9.  eval_nd_loop_5(v_0,v_x_0)      -> eval_nd_loop_bb2_in(v_0,v_x_0) [v_x_0 >= 0 && -1 + v_0 + -1*v_x_0 >= 2]                               (1,1)
          10. eval_nd_loop_5(v_0,v_x_0)      -> eval_nd_loop_bb2_in(v_0,v_x_0) [v_x_0 >= 0 && 0 >= v_0 + -1*v_x_0]                                    (1,1)
          11. eval_nd_loop_5(v_0,v_x_0)      -> eval_nd_loop_bb2_in(v_0,v_x_0) [v_x_0 >= 0 && v_0 >= 10]                                              (1,1)
          12. eval_nd_loop_bb2_in(v_0,v_x_0) -> eval_nd_loop_stop(v_0,v_x_0)   [v_x_0 >= 0]                                                           (1,1)
        Signature:
          {(eval_nd_loop_0,2)
          ;(eval_nd_loop_1,2)
          ;(eval_nd_loop_2,2)
          ;(eval_nd_loop_3,2)
          ;(eval_nd_loop_4,2)
          ;(eval_nd_loop_5,2)
          ;(eval_nd_loop_bb0_in,2)
          ;(eval_nd_loop_bb1_in,2)
          ;(eval_nd_loop_bb2_in,2)
          ;(eval_nd_loop_start,2)
          ;(eval_nd_loop_stop,2)}
        Flow Graph:
          [0->{1},1->{2},2->{3},3->{4},4->{5},5->{6},6->{7},7->{8,9,10,11},8->{6},9->{12},10->{12},11->{12},12->{}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8,9,10,11,12]
          |
          `- p:[6,8,7] c: [8]

YES