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

YES(?,O(n^1))