YES(?,O(n^1))
* Step 1: TrivialSCCs WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. evaleasy2start(A)    -> evaleasy2entryin(A)    True     (1,1)
          1. evaleasy2entryin(A)  -> evaleasy2bb1in(A)      True     (?,1)
          2. evaleasy2bb1in(A)    -> evaleasy2bbin(A)       [A >= 1] (?,1)
          3. evaleasy2bb1in(A)    -> evaleasy2returnin(A)   [0 >= A] (?,1)
          4. evaleasy2bbin(A)     -> evaleasy2bb1in(-1 + A) True     (?,1)
          5. evaleasy2returnin(A) -> evaleasy2stop(A)       True     (?,1)
        Signature:
          {(evaleasy2bb1in,1)
          ;(evaleasy2bbin,1)
          ;(evaleasy2entryin,1)
          ;(evaleasy2returnin,1)
          ;(evaleasy2start,1)
          ;(evaleasy2stop,1)}
        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. evaleasy2start(A)    -> evaleasy2entryin(A)    True     (1,1)
          1. evaleasy2entryin(A)  -> evaleasy2bb1in(A)      True     (1,1)
          2. evaleasy2bb1in(A)    -> evaleasy2bbin(A)       [A >= 1] (?,1)
          3. evaleasy2bb1in(A)    -> evaleasy2returnin(A)   [0 >= A] (1,1)
          4. evaleasy2bbin(A)     -> evaleasy2bb1in(-1 + A) True     (?,1)
          5. evaleasy2returnin(A) -> evaleasy2stop(A)       True     (1,1)
        Signature:
          {(evaleasy2bb1in,1)
          ;(evaleasy2bbin,1)
          ;(evaleasy2entryin,1)
          ;(evaleasy2returnin,1)
          ;(evaleasy2start,1)
          ;(evaleasy2stop,1)}
        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(evaleasy2bb1in) = 2 + x1
              p(evaleasy2bbin) = 1 + x1
           p(evaleasy2entryin) = 2 + x1
          p(evaleasy2returnin) = 2 + x1
             p(evaleasy2start) = 2 + x1
              p(evaleasy2stop) = 2 + x1
        
        Following rules are strictly oriented:
                   [A >= 1] ==>                 
          evaleasy2bb1in(A)   = 2 + A           
                              > 1 + A           
                              = evaleasy2bbin(A)
        
        
        Following rules are weakly oriented:
                          True ==>                       
             evaleasy2start(A)   = 2 + A                 
                                >= 2 + A                 
                                 = evaleasy2entryin(A)   
        
                          True ==>                       
           evaleasy2entryin(A)   = 2 + A                 
                                >= 2 + A                 
                                 = evaleasy2bb1in(A)     
        
                      [0 >= A] ==>                       
             evaleasy2bb1in(A)   = 2 + A                 
                                >= 2 + A                 
                                 = evaleasy2returnin(A)  
        
                          True ==>                       
              evaleasy2bbin(A)   = 1 + A                 
                                >= 1 + A                 
                                 = evaleasy2bb1in(-1 + A)
        
                          True ==>                       
          evaleasy2returnin(A)   = 2 + A                 
                                >= 2 + A                 
                                 = evaleasy2stop(A)      
        
        
* Step 3: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. evaleasy2start(A)    -> evaleasy2entryin(A)    True     (1,1)    
          1. evaleasy2entryin(A)  -> evaleasy2bb1in(A)      True     (1,1)    
          2. evaleasy2bb1in(A)    -> evaleasy2bbin(A)       [A >= 1] (2 + A,1)
          3. evaleasy2bb1in(A)    -> evaleasy2returnin(A)   [0 >= A] (1,1)    
          4. evaleasy2bbin(A)     -> evaleasy2bb1in(-1 + A) True     (?,1)    
          5. evaleasy2returnin(A) -> evaleasy2stop(A)       True     (1,1)    
        Signature:
          {(evaleasy2bb1in,1)
          ;(evaleasy2bbin,1)
          ;(evaleasy2entryin,1)
          ;(evaleasy2returnin,1)
          ;(evaleasy2start,1)
          ;(evaleasy2stop,1)}
        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. evaleasy2start(A)    -> evaleasy2entryin(A)    True     (1,1)    
          1. evaleasy2entryin(A)  -> evaleasy2bb1in(A)      True     (1,1)    
          2. evaleasy2bb1in(A)    -> evaleasy2bbin(A)       [A >= 1] (2 + A,1)
          3. evaleasy2bb1in(A)    -> evaleasy2returnin(A)   [0 >= A] (1,1)    
          4. evaleasy2bbin(A)     -> evaleasy2bb1in(-1 + A) True     (2 + A,1)
          5. evaleasy2returnin(A) -> evaleasy2stop(A)       True     (1,1)    
        Signature:
          {(evaleasy2bb1in,1)
          ;(evaleasy2bbin,1)
          ;(evaleasy2entryin,1)
          ;(evaleasy2returnin,1)
          ;(evaleasy2start,1)
          ;(evaleasy2stop,1)}
        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))