WORST_CASE(?,O(n^1))
* Step 1: NaturalMI WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            append(l1,l2) -> ifappend(l1,l2,l1)
            hd(cons(x,l)) -> x
            ifappend(l1,l2,cons(x,l)) -> cons(x,append(l,l2))
            ifappend(l1,l2,nil()) -> l2
            is_empty(cons(x,l)) -> false()
            is_empty(nil()) -> true()
            tl(cons(x,l)) -> l
        - Signature:
            {append/2,hd/1,ifappend/3,is_empty/1,tl/1} / {cons/2,false/0,nil/0,true/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {append,hd,ifappend,is_empty,tl} and constructors {cons
            ,false,nil,true}
    + Applied Processor:
        NaturalMI {miDimension = 1, miDegree = 1, miKind = Algebraic, uargs = UArgs, urules = URules, selector = Nothing}
    + Details:
        We apply a matrix interpretation of kind constructor based matrix interpretation:
        The following argument positions are considered usable:
          uargs(cons) = {2}
        
        Following symbols are considered usable:
          {append,hd,ifappend,is_empty,tl}
        TcT has computed the following interpretation:
            p(append) = [4] x1 + [1] x2 + [4]
              p(cons) = [1] x1 + [1] x2 + [1]
             p(false) = [10]                 
                p(hd) = [1] x1 + [10]        
          p(ifappend) = [1] x2 + [4] x3 + [2]
          p(is_empty) = [9] x1 + [7]         
               p(nil) = [1]                  
                p(tl) = [12] x1 + [6]        
              p(true) = [1]                  
        
        Following rules are strictly oriented:
                    append(l1,l2) = [4] l1 + [1] l2 + [4]       
                                  > [4] l1 + [1] l2 + [2]       
                                  = ifappend(l1,l2,l1)          
        
                    hd(cons(x,l)) = [1] l + [1] x + [11]        
                                  > [1] x + [0]                 
                                  = x                           
        
        ifappend(l1,l2,cons(x,l)) = [4] l + [1] l2 + [4] x + [6]
                                  > [4] l + [1] l2 + [1] x + [5]
                                  = cons(x,append(l,l2))        
        
            ifappend(l1,l2,nil()) = [1] l2 + [6]                
                                  > [1] l2 + [0]                
                                  = l2                          
        
              is_empty(cons(x,l)) = [9] l + [9] x + [16]        
                                  > [10]                        
                                  = false()                     
        
                  is_empty(nil()) = [16]                        
                                  > [1]                         
                                  = true()                      
        
                    tl(cons(x,l)) = [12] l + [12] x + [18]      
                                  > [1] l + [0]                 
                                  = l                           
        
        
        Following rules are (at-least) weakly oriented:
        

WORST_CASE(?,O(n^1))