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

YES(?,O(n^1))