WORST_CASE(?,O(n^1))
* Step 1: NaturalPI WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16)
            dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2))
            main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil())
            revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2))
            revApp#2(Nil(),x16) -> x16
        - Signature:
            {dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil
            ,Node}
    + Applied Processor:
        NaturalPI {shape = Linear, restrict = Restrict, uargs = UArgs, urules = URules, selector = Just any strict-rules}
    + Details:
        We apply a polynomial interpretation of kind constructor-based(linear):
        The following argument positions are considered usable:
          uargs(dfsAcc#3) = {2},
          uargs(revApp#2) = {1}
        
        Following symbols are considered usable:
          {dfsAcc#3,main,revApp#2}
        TcT has computed the following interpretation:
              p(Cons) = 6 + x1 + x2    
              p(Leaf) = 8 + x1         
               p(Nil) = 0              
              p(Node) = 8 + x1 + x2    
          p(dfsAcc#3) = 1 + x1 + x2    
              p(main) = 8 + 8*x1       
          p(revApp#2) = 4 + 4*x1 + 2*x2
        
        Following rules are strictly oriented:
          dfsAcc#3(Leaf(x8),x16) = 9 + x16 + x8                
                                 > 6 + x16 + x8                
                                 = Cons(x8,x16)                
        
        dfsAcc#3(Node(x6,x4),x2) = 9 + x2 + x4 + x6            
                                 > 2 + x2 + x4 + x6            
                                 = dfsAcc#3(x4,dfsAcc#3(x6,x2))
        
        revApp#2(Cons(x6,x4),x2) = 28 + 2*x2 + 4*x4 + 4*x6     
                                 > 16 + 2*x2 + 4*x4 + 2*x6     
                                 = revApp#2(x4,Cons(x6,x2))    
        
             revApp#2(Nil(),x16) = 4 + 2*x16                   
                                 > x16                         
                                 = x16                         
        
        
        Following rules are (at-least) weakly oriented:
        main(x1) =  8 + 8*x1                          
                 >= 8 + 4*x1                          
                 =  revApp#2(dfsAcc#3(x1,Nil()),Nil())
        
* Step 2: NaturalPI WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil())
        - Weak TRS:
            dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16)
            dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2))
            revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2))
            revApp#2(Nil(),x16) -> x16
        - Signature:
            {dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil
            ,Node}
    + Applied Processor:
        NaturalPI {shape = Linear, restrict = Restrict, uargs = UArgs, urules = URules, selector = Just any strict-rules}
    + Details:
        We apply a polynomial interpretation of kind constructor-based(linear):
        The following argument positions are considered usable:
          uargs(dfsAcc#3) = {2},
          uargs(revApp#2) = {1}
        
        Following symbols are considered usable:
          {dfsAcc#3,main,revApp#2}
        TcT has computed the following interpretation:
              p(Cons) = x2           
              p(Leaf) = x1           
               p(Nil) = 1            
              p(Node) = 8 + x1 + x2  
          p(dfsAcc#3) = 5 + 3*x1 + x2
              p(main) = 15 + 9*x1    
          p(revApp#2) = 2*x1 + 2*x2  
        
        Following rules are strictly oriented:
        main(x1) = 15 + 9*x1                         
                 > 14 + 6*x1                         
                 = revApp#2(dfsAcc#3(x1,Nil()),Nil())
        
        
        Following rules are (at-least) weakly oriented:
          dfsAcc#3(Leaf(x8),x16) =  5 + x16 + 3*x8              
                                 >= x16                         
                                 =  Cons(x8,x16)                
        
        dfsAcc#3(Node(x6,x4),x2) =  29 + x2 + 3*x4 + 3*x6       
                                 >= 10 + x2 + 3*x4 + 3*x6       
                                 =  dfsAcc#3(x4,dfsAcc#3(x6,x2))
        
        revApp#2(Cons(x6,x4),x2) =  2*x2 + 2*x4                 
                                 >= 2*x2 + 2*x4                 
                                 =  revApp#2(x4,Cons(x6,x2))    
        
             revApp#2(Nil(),x16) =  2 + 2*x16                   
                                 >= x16                         
                                 =  x16                         
        
* Step 3: EmptyProcessor WORST_CASE(?,O(1))
    + Considered Problem:
        - Weak TRS:
            dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16)
            dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2))
            main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil())
            revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2))
            revApp#2(Nil(),x16) -> x16
        - Signature:
            {dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {dfsAcc#3,main,revApp#2} and constructors {Cons,Leaf,Nil
            ,Node}
    + Applied Processor:
        EmptyProcessor
    + Details:
        The problem is already closed. The intended complexity is O(1).

WORST_CASE(?,O(n^1))