YES(?,O(1))
* Step 1: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. evalndloopstart(A)    -> evalndloopentryin(A)  True                                           (1,1)
          1. evalndloopentryin(A)  -> evalndloopbbin(0)     True                                           (?,1)
          2. evalndloopbbin(A)     -> evalndloopbbin(B)     [A >= 0 && 2 + A >= B && B >= 1 + A && 9 >= B] (?,1)
          3. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && B >= 3 + A]                         (?,1)
          4. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && A >= B]                             (?,1)
          5. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && B >= 10]                            (?,1)
          6. evalndloopreturnin(A) -> evalndloopstop(A)     [A >= 0]                                       (?,1)
        Signature:
          {(evalndloopbbin,1);(evalndloopentryin,1);(evalndloopreturnin,1);(evalndloopstart,1);(evalndloopstop,1)}
        Flow Graph:
          [0->{1},1->{2,3,4,5},2->{2,3,4,5},3->{6},4->{6},5->{6},6->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. evalndloopstart(A)    -> evalndloopentryin(A)  True                                           (1,1)
          1. evalndloopentryin(A)  -> evalndloopbbin(0)     True                                           (1,1)
          2. evalndloopbbin(A)     -> evalndloopbbin(B)     [A >= 0 && 2 + A >= B && B >= 1 + A && 9 >= B] (?,1)
          3. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && B >= 3 + A]                         (1,1)
          4. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && A >= B]                             (1,1)
          5. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && B >= 10]                            (1,1)
          6. evalndloopreturnin(A) -> evalndloopstop(A)     [A >= 0]                                       (1,1)
        Signature:
          {(evalndloopbbin,1);(evalndloopentryin,1);(evalndloopreturnin,1);(evalndloopstart,1);(evalndloopstop,1)}
        Flow Graph:
          [0->{1},1->{2,3,4,5},2->{2,3,4,5},3->{6},4->{6},5->{6},6->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
              p(evalndloopbbin) = 9 + -1*x1
           p(evalndloopentryin) = 9        
          p(evalndloopreturnin) = 9 + -1*x1
             p(evalndloopstart) = 9        
              p(evalndloopstop) = 9 + -1*x1
        
        Following rules are strictly oriented:
        [A >= 0 && 2 + A >= B && B >= 1 + A && 9 >= B] ==>                  
                                     evalndloopbbin(A)   = 9 + -1*A         
                                                         > 9 + -1*B         
                                                         = evalndloopbbin(B)
        
        
        Following rules are weakly oriented:
                           True ==>                      
             evalndloopstart(A)   = 9                    
                                 >= 9                    
                                  = evalndloopentryin(A) 
        
                           True ==>                      
           evalndloopentryin(A)   = 9                    
                                 >= 9                    
                                  = evalndloopbbin(0)    
        
         [A >= 0 && B >= 3 + A] ==>                      
              evalndloopbbin(A)   = 9 + -1*A             
                                 >= 9 + -1*A             
                                  = evalndloopreturnin(A)
        
             [A >= 0 && A >= B] ==>                      
              evalndloopbbin(A)   = 9 + -1*A             
                                 >= 9 + -1*A             
                                  = evalndloopreturnin(A)
        
            [A >= 0 && B >= 10] ==>                      
              evalndloopbbin(A)   = 9 + -1*A             
                                 >= 9 + -1*A             
                                  = evalndloopreturnin(A)
        
                       [A >= 0] ==>                      
          evalndloopreturnin(A)   = 9 + -1*A             
                                 >= 9 + -1*A             
                                  = evalndloopstop(A)    
        
        
* Step 3: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. evalndloopstart(A)    -> evalndloopentryin(A)  True                                           (1,1)
          1. evalndloopentryin(A)  -> evalndloopbbin(0)     True                                           (1,1)
          2. evalndloopbbin(A)     -> evalndloopbbin(B)     [A >= 0 && 2 + A >= B && B >= 1 + A && 9 >= B] (9,1)
          3. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && B >= 3 + A]                         (1,1)
          4. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && A >= B]                             (1,1)
          5. evalndloopbbin(A)     -> evalndloopreturnin(A) [A >= 0 && B >= 10]                            (1,1)
          6. evalndloopreturnin(A) -> evalndloopstop(A)     [A >= 0]                                       (1,1)
        Signature:
          {(evalndloopbbin,1);(evalndloopentryin,1);(evalndloopreturnin,1);(evalndloopstart,1);(evalndloopstop,1)}
        Flow Graph:
          [0->{1},1->{2,3,4,5},2->{2,3,4,5},3->{6},4->{6},5->{6},6->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))