reserve L for complete Scott TopLattice,
  x for Element of L,
  X, Y for Subset of L,
  V, W for Element of InclPoset sigma L,
  VV for Subset of InclPoset sigma L;

theorem :: Theorem 1.14 (1) implies (3 first conjunct) p. 107
  L is continuous implies ex B being Basis of x st for X st X in B holds
  X is open filtered
proof
  set B = { V where V is Subset of L : ex q being Element of L st V c=
  wayabove q & q<<x & x in V & V is open filtered };
  B c= bool the carrier of L
  proof
    let X be object;
    assume X in B;
    then ex V being Subset of L st X = V & ex q being Element of L st V c=
    wayabove q & q<<x & x in V & V is open filtered;
    hence thesis;
  end;
  then reconsider B as Subset-Family of L;
  reconsider B as Subset-Family of L;
  assume
A1: L is continuous;
  then reconsider
  A = { wayabove q where q is Element of L: q << x } as Basis of x
  by WAYBEL11:44;
A2: B is Basis of x
  proof
A3: B is open
    proof
      let Y be Subset of L;
      assume Y in B;
      then ex V being Subset of L st Y = V & ex q being Element of L st V c=
      wayabove q & q<<x & x in V & V is open filtered;
      hence thesis;
    end;
    B is x-quasi_basis
    proof
    thus x in Intersect B
    proof
      per cases;
      suppose
        B is empty;
        then Intersect B = the carrier of L by SETFAM_1:def 9;
        hence thesis;
      end;
      suppose
A4:     B is non empty;
A5:     now
          let Y be set;
          assume Y in B;
          then ex V being Subset of L st Y = V & ex q being Element of L st V
          c= wayabove q & q<<x & x in V& V is open filtered;
          hence x in Y;
        end;
        Intersect B = meet B by A4,SETFAM_1:def 9;
        hence thesis by A4,A5,SETFAM_1:def 1;
      end;
    end;
    let S be Subset of L;
    assume S is open & x in S;
    then consider V being Subset of L such that
A6: V in A and
A7: V c= S by YELLOW_8:def 1;
    consider q being Element of L such that
A8: V = wayabove q and
A9: q << x by A6;
    consider F being Open Filter of L such that
A10: x in F and
A11: F c= wayabove q by A1,A9,WAYBEL_6:8;
    take F;
    F is open by WAYBEL11:41;
    hence F in B by A9,A10,A11;
    thus thesis by A7,A8,A11;
  end;
  hence thesis by A3;
  end;
  now
    let Y be Subset of L;
    assume Y in B;
    then ex V being Subset of L st Y = V & ex q being Element of L st V c=
    wayabove q & q<<x & x in V & V is open filtered;
    hence Y is open filtered;
  end;
  hence thesis by A2;
end;
