YES(?,O(n^1))
* Step 1: TrivialSCCs WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. evalndecrstart(A)    -> evalndecrentryin(A)    True     (1,1)
          1. evalndecrentryin(A)  -> evalndecrbb1in(-1 + A) True     (?,1)
          2. evalndecrbb1in(A)    -> evalndecrbbin(A)       [A >= 2] (?,1)
          3. evalndecrbb1in(A)    -> evalndecrreturnin(A)   [1 >= A] (?,1)
          4. evalndecrbbin(A)     -> evalndecrbb1in(-1 + A) True     (?,1)
          5. evalndecrreturnin(A) -> evalndecrstop(A)       True     (?,1)
        Signature:
          {(evalndecrbb1in,1)
          ;(evalndecrbbin,1)
          ;(evalndecrentryin,1)
          ;(evalndecrreturnin,1)
          ;(evalndecrstart,1)
          ;(evalndecrstop,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. evalndecrstart(A)    -> evalndecrentryin(A)    True     (1,1)
          1. evalndecrentryin(A)  -> evalndecrbb1in(-1 + A) True     (1,1)
          2. evalndecrbb1in(A)    -> evalndecrbbin(A)       [A >= 2] (?,1)
          3. evalndecrbb1in(A)    -> evalndecrreturnin(A)   [1 >= A] (1,1)
          4. evalndecrbbin(A)     -> evalndecrbb1in(-1 + A) True     (?,1)
          5. evalndecrreturnin(A) -> evalndecrstop(A)       True     (1,1)
        Signature:
          {(evalndecrbb1in,1)
          ;(evalndecrbbin,1)
          ;(evalndecrentryin,1)
          ;(evalndecrreturnin,1)
          ;(evalndecrstart,1)
          ;(evalndecrstop,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(evalndecrbb1in) = 2 + x1
              p(evalndecrbbin) = 1 + x1
           p(evalndecrentryin) = 1 + x1
          p(evalndecrreturnin) = 2 + x1
             p(evalndecrstart) = 1 + x1
              p(evalndecrstop) = 2 + x1
        
        Following rules are strictly oriented:
                   [A >= 2] ==>                 
          evalndecrbb1in(A)   = 2 + A           
                              > 1 + A           
                              = evalndecrbbin(A)
        
        
        Following rules are weakly oriented:
                          True ==>                       
             evalndecrstart(A)   = 1 + A                 
                                >= 1 + A                 
                                 = evalndecrentryin(A)   
        
                          True ==>                       
           evalndecrentryin(A)   = 1 + A                 
                                >= 1 + A                 
                                 = evalndecrbb1in(-1 + A)
        
                      [1 >= A] ==>                       
             evalndecrbb1in(A)   = 2 + A                 
                                >= 2 + A                 
                                 = evalndecrreturnin(A)  
        
                          True ==>                       
              evalndecrbbin(A)   = 1 + A                 
                                >= 1 + A                 
                                 = evalndecrbb1in(-1 + A)
        
                          True ==>                       
          evalndecrreturnin(A)   = 2 + A                 
                                >= 2 + A                 
                                 = evalndecrstop(A)      
        
        
* Step 3: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. evalndecrstart(A)    -> evalndecrentryin(A)    True     (1,1)    
          1. evalndecrentryin(A)  -> evalndecrbb1in(-1 + A) True     (1,1)    
          2. evalndecrbb1in(A)    -> evalndecrbbin(A)       [A >= 2] (1 + A,1)
          3. evalndecrbb1in(A)    -> evalndecrreturnin(A)   [1 >= A] (1,1)    
          4. evalndecrbbin(A)     -> evalndecrbb1in(-1 + A) True     (?,1)    
          5. evalndecrreturnin(A) -> evalndecrstop(A)       True     (1,1)    
        Signature:
          {(evalndecrbb1in,1)
          ;(evalndecrbbin,1)
          ;(evalndecrentryin,1)
          ;(evalndecrreturnin,1)
          ;(evalndecrstart,1)
          ;(evalndecrstop,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. evalndecrstart(A)    -> evalndecrentryin(A)    True     (1,1)    
          1. evalndecrentryin(A)  -> evalndecrbb1in(-1 + A) True     (1,1)    
          2. evalndecrbb1in(A)    -> evalndecrbbin(A)       [A >= 2] (1 + A,1)
          3. evalndecrbb1in(A)    -> evalndecrreturnin(A)   [1 >= A] (1,1)    
          4. evalndecrbbin(A)     -> evalndecrbb1in(-1 + A) True     (1 + A,1)
          5. evalndecrreturnin(A) -> evalndecrstop(A)       True     (1,1)    
        Signature:
          {(evalndecrbb1in,1)
          ;(evalndecrbbin,1)
          ;(evalndecrentryin,1)
          ;(evalndecrreturnin,1)
          ;(evalndecrstart,1)
          ;(evalndecrstop,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))