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

YES(?,O(n^1))