
theorem Th13:
  for X, x being set st x in X holds (disjoin Card id X).x = [: card x, {x} :]
proof
  let X, x be set;
  assume A1: x in X;
  then A2: x in dom id X;
  then x in dom Card id X by CARD_3:def 2;
  hence (disjoin Card id X).x = [: (Card id X).x, {x} :] by CARD_3:def 3
    .= [: card ((id X).x), {x} :] by A2, CARD_3:def 2
    .= [: card x, {x} :] by A1, FUNCT_1:18;
end;
