WORST_CASE(?,O(n^1))
* Step 1: NaturalMI WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            fold(a,xs) -> Cons(foldl(a,xs),Cons(foldr(a,xs),Nil()))
            foldl(a,Nil()) -> a
            foldl(x,Cons(S(0()),xs)) -> foldl(S(x),xs)
            foldl(S(0()),Cons(x,xs)) -> foldl(S(x),xs)
            foldr(a,Cons(x,xs)) -> op(x,foldr(a,xs))
            foldr(a,Nil()) -> a
            notEmpty(Cons(x,xs)) -> True()
            notEmpty(Nil()) -> False()
            op(x,S(0())) -> S(x)
            op(S(0()),y) -> S(y)
        - Signature:
            {fold/2,foldl/2,foldr/2,notEmpty/1,op/2} / {0/0,Cons/2,False/0,Nil/0,S/1,True/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {fold,foldl,foldr,notEmpty,op} and constructors {0,Cons
            ,False,Nil,S,True}
    + Applied Processor:
        NaturalMI {miDimension = 1, miDegree = 1, miKind = Algebraic, uargs = UArgs, urules = URules, selector = Just any strict-rules}
    + Details:
        We apply a matrix interpretation of kind constructor based matrix interpretation:
        The following argument positions are considered usable:
          uargs(Cons) = {1,2},
          uargs(op) = {2}
        
        Following symbols are considered usable:
          {fold,foldl,foldr,notEmpty,op}
        TcT has computed the following interpretation:
                 p(0) = [0]                   
              p(Cons) = [1] x1 + [1] x2 + [2] 
             p(False) = [1]                   
               p(Nil) = [1]                   
                 p(S) = [1] x1 + [6]          
              p(True) = [0]                   
              p(fold) = [5] x1 + [9] x2 + [14]
             p(foldl) = [2] x1 + [2] x2 + [0] 
             p(foldr) = [2] x1 + [7] x2 + [5] 
          p(notEmpty) = [2] x1 + [6]          
                p(op) = [2] x1 + [1] x2 + [5] 
        
        Following rules are strictly oriented:
                      fold(a,xs) = [5] a + [9] xs + [14]                    
                                 > [4] a + [9] xs + [10]                    
                                 = Cons(foldl(a,xs),Cons(foldr(a,xs),Nil()))
        
                  foldl(a,Nil()) = [2] a + [2]                              
                                 > [1] a + [0]                              
                                 = a                                        
        
        foldl(x,Cons(S(0()),xs)) = [2] x + [2] xs + [16]                    
                                 > [2] x + [2] xs + [12]                    
                                 = foldl(S(x),xs)                           
        
        foldl(S(0()),Cons(x,xs)) = [2] x + [2] xs + [16]                    
                                 > [2] x + [2] xs + [12]                    
                                 = foldl(S(x),xs)                           
        
             foldr(a,Cons(x,xs)) = [2] a + [7] x + [7] xs + [19]            
                                 > [2] a + [2] x + [7] xs + [10]            
                                 = op(x,foldr(a,xs))                        
        
                  foldr(a,Nil()) = [2] a + [12]                             
                                 > [1] a + [0]                              
                                 = a                                        
        
            notEmpty(Cons(x,xs)) = [2] x + [2] xs + [10]                    
                                 > [0]                                      
                                 = True()                                   
        
                 notEmpty(Nil()) = [8]                                      
                                 > [1]                                      
                                 = False()                                  
        
                    op(x,S(0())) = [2] x + [11]                             
                                 > [1] x + [6]                              
                                 = S(x)                                     
        
                    op(S(0()),y) = [1] y + [17]                             
                                 > [1] y + [6]                              
                                 = S(y)                                     
        
        
        Following rules are (at-least) weakly oriented:
        
* Step 2: EmptyProcessor WORST_CASE(?,O(1))
    + Considered Problem:
        - Weak TRS:
            fold(a,xs) -> Cons(foldl(a,xs),Cons(foldr(a,xs),Nil()))
            foldl(a,Nil()) -> a
            foldl(x,Cons(S(0()),xs)) -> foldl(S(x),xs)
            foldl(S(0()),Cons(x,xs)) -> foldl(S(x),xs)
            foldr(a,Cons(x,xs)) -> op(x,foldr(a,xs))
            foldr(a,Nil()) -> a
            notEmpty(Cons(x,xs)) -> True()
            notEmpty(Nil()) -> False()
            op(x,S(0())) -> S(x)
            op(S(0()),y) -> S(y)
        - Signature:
            {fold/2,foldl/2,foldr/2,notEmpty/1,op/2} / {0/0,Cons/2,False/0,Nil/0,S/1,True/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {fold,foldl,foldr,notEmpty,op} and constructors {0,Cons
            ,False,Nil,S,True}
    + Applied Processor:
        EmptyProcessor
    + Details:
        The problem is already closed. The intended complexity is O(1).

WORST_CASE(?,O(n^1))