reserve x, X, Y for set;

theorem
  for L being non empty RelStr for S, T being non empty full SubRelStr
of L st the carrier of S c= the carrier of T for X being Subset of S holds X is
  Subset of T & for Y being Subset of T st X = Y holds (X is filtered implies Y
  is filtered) & (X is directed implies Y is directed)
proof
  let L be non empty RelStr, S1,S2 be non empty full SubRelStr of L;
  assume
A1: the carrier of S1 c= the carrier of S2;
  let X be Subset of S1;
  thus X is Subset of S2 by A1,XBOOLE_1:1;
  let X2 be Subset of S2 such that
A2: X = X2;
  hereby
    assume
A3: X is filtered;
    thus X2 is filtered
    proof
      let x, y be Element of S2;
      assume
A4:   x in X2 & y in X2;
      then reconsider x9= x, y9= y as Element of S1 by A2;
      consider z being Element of S1 such that
A5:   z in X and
A6:   z <= x9 & z <= y9 by A2,A3,A4;
      reconsider x1 = x, y1 = y, z1 = z as Element of L by YELLOW_0:58;
      reconsider z as Element of S2 by A2,A5;
      take z;
      z1 <= x1 & z1 <= y1 by A6,YELLOW_0:59;
      hence thesis by A2,A5,YELLOW_0:60;
    end;
  end;
  assume
A7: X is directed;
  let x, y be Element of S2;
  assume
A8: x in X2 & y in X2;
  then reconsider x9= x, y9= y as Element of S1 by A2;
  consider z being Element of S1 such that
A9: z in X and
A10: x9 <= z & y9 <= z by A2,A7,A8;
  reconsider x1 = x, y1 = y, z1 = z as Element of L by YELLOW_0:58;
  reconsider z as Element of S2 by A2,A9;
  take z;
  x1 <= z1 & y1 <= z1 by A10,YELLOW_0:59;
  hence thesis by A2,A9,YELLOW_0:60;
end;
