WORST_CASE(?,O(n^1))
* Step 1: NaturalPI 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:
        NaturalPI {shape = Linear, restrict = NoRestrict, uargs = UArgs, urules = URules, selector = Nothing}
    + Details:
        We apply a polynomial interpretation of kind constructor-based(linear):
        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) = 2 + 8*x1 + 8*x2
              p(cons) = 2 + x1 + x2    
             p(false) = 15             
                p(hd) = 1 + 6*x1       
          p(ifappend) = 8*x2 + 8*x3    
          p(is_empty) = 15 + x1        
               p(nil) = 2              
                p(tl) = 10 + 4*x1      
              p(true) = 13             
        
        Following rules are strictly oriented:
                    append(l1,l2) = 2 + 8*l1 + 8*l2      
                                  > 8*l1 + 8*l2          
                                  = ifappend(l1,l2,l1)   
        
                    hd(cons(x,l)) = 13 + 6*l + 6*x       
                                  > x                    
                                  = x                    
        
        ifappend(l1,l2,cons(x,l)) = 16 + 8*l + 8*l2 + 8*x
                                  > 4 + 8*l + 8*l2 + x   
                                  = cons(x,append(l,l2)) 
        
            ifappend(l1,l2,nil()) = 16 + 8*l2            
                                  > l2                   
                                  = l2                   
        
              is_empty(cons(x,l)) = 17 + l + x           
                                  > 15                   
                                  = false()              
        
                  is_empty(nil()) = 17                   
                                  > 13                   
                                  = true()               
        
                    tl(cons(x,l)) = 18 + 4*l + 4*x       
                                  > l                    
                                  = l                    
        
        
        Following rules are (at-least) weakly oriented:
        

WORST_CASE(?,O(n^1))