reserve            x for object,
               X,Y,Z for set,
         i,j,k,l,m,n for Nat,
                 r,s for Real,
                  no for Element of OrderedNAT,
                   A for Subset of [:NAT,NAT:];

theorem Th18:
  for X being non empty set, cF being Filter of X holds
  ex cB being filter_base of X st cB = cF & <.cB.) = cF
  proof
    let X be non empty set, cF be Filter of X;
    cF is basis of cF by CARDFIL2:13;
    then reconsider cB = cF as filter_base of X by CARDFIL2:29;
    take cB;
    now
      hereby
        let x be object;
        assume
A1:     x in <.cB.);
        then reconsider y = x as Subset of X;
        ex b be Element of cB st b c= y by A1,CARDFIL2:def 8;
        hence x in cF by CARD_FIL:def 1;
      end;
      let x be object;
      assume x in cF;
      hence x in <.cB.) by CARDFIL2:def 8;
    end;
    then <.cB.) c= cF & cF c= <.cB.);
    hence thesis;
  end;
