WORST_CASE(?,O(n^1))
* Step 1: NaturalPI WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            add(0(),X) -> X
            add(s(),Y) -> s()
            dbl(0()) -> 0()
            dbl(s()) -> s()
            first(0(),X) -> nil()
            first(s(),cons(Y)) -> cons(Y)
            sqr(0()) -> 0()
            sqr(s()) -> s()
            terms(N) -> cons(recip(sqr(N)))
        - Signature:
            {add/2,dbl/1,first/2,sqr/1,terms/1} / {0/0,cons/1,nil/0,recip/1,s/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {add,dbl,first,sqr,terms} and constructors {0,cons,nil
            ,recip,s}
    + 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) = {1},
          uargs(recip) = {1}
        
        Following symbols are considered usable:
          {add,dbl,first,sqr,terms}
        TcT has computed the following interpretation:
              p(0) = 0               
            p(add) = 11 + 8*x2       
           p(cons) = x1              
            p(dbl) = 4 + 2*x1        
          p(first) = 10 + 2*x1 + 8*x2
            p(nil) = 7               
          p(recip) = 4 + x1          
              p(s) = 3               
            p(sqr) = 8               
          p(terms) = 14              
        
        Following rules are strictly oriented:
                add(0(),X) = 11 + 8*X           
                           > X                  
                           = X                  
        
                add(s(),Y) = 11 + 8*Y           
                           > 3                  
                           = s()                
        
                  dbl(0()) = 4                  
                           > 0                  
                           = 0()                
        
                  dbl(s()) = 10                 
                           > 3                  
                           = s()                
        
              first(0(),X) = 10 + 8*X           
                           > 7                  
                           = nil()              
        
        first(s(),cons(Y)) = 16 + 8*Y           
                           > Y                  
                           = cons(Y)            
        
                  sqr(0()) = 8                  
                           > 0                  
                           = 0()                
        
                  sqr(s()) = 8                  
                           > 3                  
                           = s()                
        
                  terms(N) = 14                 
                           > 12                 
                           = cons(recip(sqr(N)))
        
        
        Following rules are (at-least) weakly oriented:
        

WORST_CASE(?,O(n^1))