YES(?,O(n^1))
* Step 1: UnsatPaths WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B) -> eval(A,B)      True                                 (1,1)
          1. eval(A,B)  -> eval(A,-1 + B) [A + B >= 1 && B >= A && B >= 1 + A] (?,1)
          2. eval(A,B)  -> eval(-1 + A,B) [2*A >= 1 && B = A]                  (?,1)
          3. eval(A,B)  -> eval(-1 + A,B) [A + B >= 1 && A >= 1 + B]           (?,1)
        Signature:
          {(eval,2);(start,2)}
        Flow Graph:
          [0->{1,2,3},1->{1,2,3},2->{1,2,3},3->{1,2,3}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(1,3),(2,2),(2,3),(3,1)]
* Step 2: PolyRank WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B) -> eval(A,B)      True                                 (1,1)
          1. eval(A,B)  -> eval(A,-1 + B) [A + B >= 1 && B >= A && B >= 1 + A] (?,1)
          2. eval(A,B)  -> eval(-1 + A,B) [2*A >= 1 && B = A]                  (?,1)
          3. eval(A,B)  -> eval(-1 + A,B) [A + B >= 1 && A >= 1 + B]           (?,1)
        Signature:
          {(eval,2);(start,2)}
        Flow Graph:
          [0->{1,2,3},1->{1,2},2->{1},3->{2,3}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(eval) = 2*x1
          p(start) = 2*x1
        
        Following rules are strictly oriented:
        [A + B >= 1 && A >= 1 + B] ==>               
                         eval(A,B)   = 2*A           
                                     > -2 + 2*A      
                                     = eval(-1 + A,B)
        
        
        Following rules are weakly oriented:
                                        True ==>               
                                  start(A,B)   = 2*A           
                                              >= 2*A           
                                               = eval(A,B)     
        
        [A + B >= 1 && B >= A && B >= 1 + A] ==>               
                                   eval(A,B)   = 2*A           
                                              >= 2*A           
                                               = eval(A,-1 + B)
        
                         [2*A >= 1 && B = A] ==>               
                                   eval(A,B)   = 2*A           
                                              >= -2 + 2*A      
                                               = eval(-1 + A,B)
        
        
* Step 3: PolyRank WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B) -> eval(A,B)      True                                 (1,1)  
          1. eval(A,B)  -> eval(A,-1 + B) [A + B >= 1 && B >= A && B >= 1 + A] (?,1)  
          2. eval(A,B)  -> eval(-1 + A,B) [2*A >= 1 && B = A]                  (?,1)  
          3. eval(A,B)  -> eval(-1 + A,B) [A + B >= 1 && A >= 1 + B]           (2*A,1)
        Signature:
          {(eval,2);(start,2)}
        Flow Graph:
          [0->{1,2,3},1->{1,2},2->{1},3->{2,3}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(eval) = 2*x1
          p(start) = 2*x1
        
        Following rules are strictly oriented:
               [2*A >= 1 && B = A] ==>               
                         eval(A,B)   = 2*A           
                                     > -2 + 2*A      
                                     = eval(-1 + A,B)
        
        [A + B >= 1 && A >= 1 + B] ==>               
                         eval(A,B)   = 2*A           
                                     > -2 + 2*A      
                                     = eval(-1 + A,B)
        
        
        Following rules are weakly oriented:
                                        True ==>               
                                  start(A,B)   = 2*A           
                                              >= 2*A           
                                               = eval(A,B)     
        
        [A + B >= 1 && B >= A && B >= 1 + A] ==>               
                                   eval(A,B)   = 2*A           
                                              >= 2*A           
                                               = eval(A,-1 + B)
        
        
* Step 4: PolyRank WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B) -> eval(A,B)      True                                 (1,1)  
          1. eval(A,B)  -> eval(A,-1 + B) [A + B >= 1 && B >= A && B >= 1 + A] (?,1)  
          2. eval(A,B)  -> eval(-1 + A,B) [2*A >= 1 && B = A]                  (2*A,1)
          3. eval(A,B)  -> eval(-1 + A,B) [A + B >= 1 && A >= 1 + B]           (2*A,1)
        Signature:
          {(eval,2);(start,2)}
        Flow Graph:
          [0->{1,2,3},1->{1,2},2->{1},3->{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:
        [A + B >= 1 && B >= A && B >= 1 + A] ==>               
                                   eval(A,B)   = 2*B           
                                               > -2 + 2*B      
                                               = eval(A,-1 + B)
        
        
        Following rules are weakly oriented:
                              True ==>               
                        start(A,B)   = 2*B           
                                    >= 2*B           
                                     = eval(A,B)     
        
               [2*A >= 1 && B = A] ==>               
                         eval(A,B)   = 2*B           
                                    >= 2*B           
                                     = eval(-1 + A,B)
        
        [A + B >= 1 && A >= 1 + B] ==>               
                         eval(A,B)   = 2*B           
                                    >= 2*B           
                                     = eval(-1 + A,B)
        
        
* Step 5: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B) -> eval(A,B)      True                                 (1,1)  
          1. eval(A,B)  -> eval(A,-1 + B) [A + B >= 1 && B >= A && B >= 1 + A] (2*B,1)
          2. eval(A,B)  -> eval(-1 + A,B) [2*A >= 1 && B = A]                  (2*A,1)
          3. eval(A,B)  -> eval(-1 + A,B) [A + B >= 1 && A >= 1 + B]           (2*A,1)
        Signature:
          {(eval,2);(start,2)}
        Flow Graph:
          [0->{1,2,3},1->{1,2},2->{1},3->{2,3}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(n^1))