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) [-1 + A >= 0] (?,1)
          5. evaleasy2returnin(A) -> evaleasy2stop(A)       [-1*A >= 0]   (?,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) [-1 + A >= 0] (?,1)
          5. evaleasy2returnin(A) -> evaleasy2stop(A)       [-1*A >= 0]   (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) = x1
              p(evaleasy2bbin) = x1
           p(evaleasy2entryin) = x1
          p(evaleasy2returnin) = x1
             p(evaleasy2start) = x1
              p(evaleasy2stop) = x1
        
        Following rules are strictly oriented:
             [-1 + A >= 0] ==>                       
          evaleasy2bbin(A)   = A                     
                             > -1 + A                
                             = evaleasy2bb1in(-1 + A)
        
        
        Following rules are weakly oriented:
                          True ==>                     
             evaleasy2start(A)   = A                   
                                >= A                   
                                 = evaleasy2entryin(A) 
        
                          True ==>                     
           evaleasy2entryin(A)   = A                   
                                >= A                   
                                 = evaleasy2bb1in(A)   
        
                      [A >= 1] ==>                     
             evaleasy2bb1in(A)   = A                   
                                >= A                   
                                 = evaleasy2bbin(A)    
        
                      [0 >= A] ==>                     
             evaleasy2bb1in(A)   = A                   
                                >= A                   
                                 = evaleasy2returnin(A)
        
                   [-1*A >= 0] ==>                     
          evaleasy2returnin(A)   = A                   
                                >= 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]      (?,1)
          3. evaleasy2bb1in(A)    -> evaleasy2returnin(A)   [0 >= A]      (1,1)
          4. evaleasy2bbin(A)     -> evaleasy2bb1in(-1 + A) [-1 + A >= 0] (A,1)
          5. evaleasy2returnin(A) -> evaleasy2stop(A)       [-1*A >= 0]   (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]      (1 + A,1)
          3. evaleasy2bb1in(A)    -> evaleasy2returnin(A)   [0 >= A]      (1,1)    
          4. evaleasy2bbin(A)     -> evaleasy2bb1in(-1 + A) [-1 + A >= 0] (A,1)    
          5. evaleasy2returnin(A) -> evaleasy2stop(A)       [-1*A >= 0]   (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))