YES(?,O(n^1))
* Step 1: TrivialSCCs WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. evalSimpleSinglestart(A,B)    -> evalSimpleSingleentryin(A,B)   True                                                             (1,1)
          1. evalSimpleSingleentryin(A,B)  -> evalSimpleSinglebb3in(0,B)     True                                                             (?,1)
          2. evalSimpleSinglebb3in(A,B)    -> evalSimpleSinglebbin(A,B)      [A >= 0 && B >= 1 + A]                                           (?,1)
          3. evalSimpleSinglebb3in(A,B)    -> evalSimpleSinglereturnin(A,B)  [A >= 0 && A >= B]                                               (?,1)
          4. evalSimpleSinglebbin(A,B)     -> evalSimpleSinglebb3in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1)
          5. evalSimpleSinglereturnin(A,B) -> evalSimpleSinglestop(A,B)      [A + -1*B >= 0 && A >= 0]                                        (?,1)
        Signature:
          {(evalSimpleSinglebb3in,2)
          ;(evalSimpleSinglebbin,2)
          ;(evalSimpleSingleentryin,2)
          ;(evalSimpleSinglereturnin,2)
          ;(evalSimpleSinglestart,2)
          ;(evalSimpleSinglestop,2)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: PolyRank WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. evalSimpleSinglestart(A,B)    -> evalSimpleSingleentryin(A,B)   True                                                             (1,1)
          1. evalSimpleSingleentryin(A,B)  -> evalSimpleSinglebb3in(0,B)     True                                                             (1,1)
          2. evalSimpleSinglebb3in(A,B)    -> evalSimpleSinglebbin(A,B)      [A >= 0 && B >= 1 + A]                                           (?,1)
          3. evalSimpleSinglebb3in(A,B)    -> evalSimpleSinglereturnin(A,B)  [A >= 0 && A >= B]                                               (1,1)
          4. evalSimpleSinglebbin(A,B)     -> evalSimpleSinglebb3in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1)
          5. evalSimpleSinglereturnin(A,B) -> evalSimpleSinglestop(A,B)      [A + -1*B >= 0 && A >= 0]                                        (1,1)
        Signature:
          {(evalSimpleSinglebb3in,2)
          ;(evalSimpleSinglebbin,2)
          ;(evalSimpleSingleentryin,2)
          ;(evalSimpleSinglereturnin,2)
          ;(evalSimpleSinglestart,2)
          ;(evalSimpleSinglestop,2)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
             p(evalSimpleSinglebb3in) = -1*x1 + x2
              p(evalSimpleSinglebbin) = -1*x1 + x2
           p(evalSimpleSingleentryin) = x2        
          p(evalSimpleSinglereturnin) = -1*x1 + x2
             p(evalSimpleSinglestart) = x2        
              p(evalSimpleSinglestop) = -1*x1 + x2
        
        Following rules are strictly oriented:
        [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] ==>                               
                                               evalSimpleSinglebbin(A,B)   = -1*A + B                      
                                                                           > -1 + -1*A + B                 
                                                                           = evalSimpleSinglebb3in(1 + A,B)
        
        
        Following rules are weakly oriented:
                                   True ==>                              
             evalSimpleSinglestart(A,B)   = B                            
                                         >= B                            
                                          = evalSimpleSingleentryin(A,B) 
        
                                   True ==>                              
           evalSimpleSingleentryin(A,B)   = B                            
                                         >= B                            
                                          = evalSimpleSinglebb3in(0,B)   
        
                 [A >= 0 && B >= 1 + A] ==>                              
             evalSimpleSinglebb3in(A,B)   = -1*A + B                     
                                         >= -1*A + B                     
                                          = evalSimpleSinglebbin(A,B)    
        
                     [A >= 0 && A >= B] ==>                              
             evalSimpleSinglebb3in(A,B)   = -1*A + B                     
                                         >= -1*A + B                     
                                          = evalSimpleSinglereturnin(A,B)
        
              [A + -1*B >= 0 && A >= 0] ==>                              
          evalSimpleSinglereturnin(A,B)   = -1*A + B                     
                                         >= -1*A + B                     
                                          = evalSimpleSinglestop(A,B)    
        
        
* Step 3: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. evalSimpleSinglestart(A,B)    -> evalSimpleSingleentryin(A,B)   True                                                             (1,1)
          1. evalSimpleSingleentryin(A,B)  -> evalSimpleSinglebb3in(0,B)     True                                                             (1,1)
          2. evalSimpleSinglebb3in(A,B)    -> evalSimpleSinglebbin(A,B)      [A >= 0 && B >= 1 + A]                                           (?,1)
          3. evalSimpleSinglebb3in(A,B)    -> evalSimpleSinglereturnin(A,B)  [A >= 0 && A >= B]                                               (1,1)
          4. evalSimpleSinglebbin(A,B)     -> evalSimpleSinglebb3in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (B,1)
          5. evalSimpleSinglereturnin(A,B) -> evalSimpleSinglestop(A,B)      [A + -1*B >= 0 && A >= 0]                                        (1,1)
        Signature:
          {(evalSimpleSinglebb3in,2)
          ;(evalSimpleSinglebbin,2)
          ;(evalSimpleSingleentryin,2)
          ;(evalSimpleSinglereturnin,2)
          ;(evalSimpleSinglestart,2)
          ;(evalSimpleSinglestop,2)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        We propagate bounds from predecessors.
* Step 4: PolyRank WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. evalSimpleSinglestart(A,B)    -> evalSimpleSingleentryin(A,B)   True                                                             (1,1)    
          1. evalSimpleSingleentryin(A,B)  -> evalSimpleSinglebb3in(0,B)     True                                                             (1,1)    
          2. evalSimpleSinglebb3in(A,B)    -> evalSimpleSinglebbin(A,B)      [A >= 0 && B >= 1 + A]                                           (1 + B,1)
          3. evalSimpleSinglebb3in(A,B)    -> evalSimpleSinglereturnin(A,B)  [A >= 0 && A >= B]                                               (1,1)    
          4. evalSimpleSinglebbin(A,B)     -> evalSimpleSinglebb3in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (B,1)    
          5. evalSimpleSinglereturnin(A,B) -> evalSimpleSinglestop(A,B)      [A + -1*B >= 0 && A >= 0]                                        (1,1)    
        Signature:
          {(evalSimpleSinglebb3in,2)
          ;(evalSimpleSinglebbin,2)
          ;(evalSimpleSingleentryin,2)
          ;(evalSimpleSinglereturnin,2)
          ;(evalSimpleSinglestart,2)
          ;(evalSimpleSinglestop,2)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        The problem is already solved.

YES(?,O(n^1))