
theorem Th23:
  for L being RelStr, S being full SubRelStr of L for X being
  Subset of S holds (X is directed Subset of L implies X is directed) & (X is
  filtered Subset of L implies X is filtered)
proof
  let L be RelStr, S be full SubRelStr of L;
  let X be Subset of S;
  hereby
    assume
A1: X is directed Subset of L;
    thus X is directed
    proof
A2:   the carrier of S c= the carrier of L by YELLOW_0:def 13;
      let x,y be Element of S;
      assume that
A3:   x in X and
A4:   y in X;
      x in the carrier of S by A3;
      then reconsider x9 = x, y9 = y as Element of L by A2;
      consider z being Element of L such that
A5:   z in X and
A6:   z >= x9 and
A7:   z >= y9 by A1,A3,A4,WAYBEL_0:def 1;
      reconsider z as Element of S by A5;
      take z;
      thus thesis by A5,A6,A7,YELLOW_0:60;
    end;
  end;
  assume
A8: X is filtered Subset of L;
A9: the carrier of S c= the carrier of L by YELLOW_0:def 13;
  let x,y be Element of S;
  assume that
A10: x in X and
A11: y in X;
  x in the carrier of S by A10;
  then reconsider x9 = x, y9 = y as Element of L by A9;
  consider z being Element of L such that
A12: z in X and
A13: z <= x9 and
A14: z <= y9 by A8,A10,A11,WAYBEL_0:def 2;
  reconsider z as Element of S by A12;
  take z;
  thus thesis by A12,A13,A14,YELLOW_0:60;
end;
