
theorem
  for R being non empty RelStr, X being non empty Subset of R holds
  the carrier of X opp+id = X & X opp+id is full SubRelStr of R opp &
  for x being Element of X opp+id holds X opp+id.x = x
proof
  let R be non empty RelStr, X be non empty Subset of R;
A1: the RelStr of X opp+id = the RelStr of (subrelstr X)opp+id by
WAYBEL_9:def 8;
A2: the mapping of X opp+id = incl(subrelstr X, R)*the mapping of (subrelstr
  X) opp+id by WAYBEL_9:def 8;
A3: the carrier of subrelstr X = X by YELLOW_0:def 15;
A4: the carrier of (subrelstr X) opp+id = the carrier of subrelstr X by
WAYBEL_9:def 6;
A5: the InternalRel of (subrelstr X) opp+id = (the InternalRel of subrelstr
  X)~ by WAYBEL_9:def 6;
  thus
  the carrier of X opp+id = X by A1,A3,WAYBEL_9:def 6;
A6: R opp = RelStr(#the carrier of R, (the InternalRel of R)~#)
  by LATTICE3:def 5;
  the InternalRel of subrelstr X = (the InternalRel of R)|_2 X
  by A3,YELLOW_0:def 14;
  then
A7: the InternalRel of (subrelstr X)opp+id = (the InternalRel of R)~|_2 X
  by A5,ORDERS_1:83;
  then the InternalRel of (subrelstr X)opp+id c= the InternalRel of R opp
  by A6,XBOOLE_1:17;
  then reconsider S = X opp+id as SubRelStr of R opp by A1,A3,A4,A6,
YELLOW_0:def 13;
  the InternalRel of S
  = (the InternalRel of R opp)|_2 the carrier of S by A1,A4,A6,A7,
YELLOW_0:def 15;
  hence X opp+id is full SubRelStr of R opp by YELLOW_0:def 14;
  let x be Element of X opp+id;
  id subrelstr X = id X by YELLOW_0:def 15;
  then
A8: the mapping of (subrelstr X)opp+id = id X by WAYBEL_9:def 6;
A9: dom id X = X;
  incl(subrelstr X, R) = id X by A3,Def1;
  then the mapping of X opp+id = id X by A2,A8,A9,RELAT_1:52;
  hence thesis by A1,A3,A4;
end;
