
theorem Th27:
  for f being Function, X, Y being set, i being object st X c= Y
  holds product(f +* (i .--> X)) c= product(f +* (i .--> Y))
proof
  let f be Function, X, Y be set, i be object;
  assume A1: X c= Y;
  dom (f +* (i .--> X)) = dom f \/ dom(i .--> X) by FUNCT_4:def 1
    .= dom f \/ dom({i} --> X) by FUNCOP_1:def 9
    .= dom f \/ {i};
  then A3: dom (f +* (i .--> X)) = dom f \/ dom({i} --> Y)
    .= dom f \/ dom(i .--> Y) by FUNCOP_1:def 9
    .= dom (f +* (i .--> Y)) by FUNCT_4:def 1;
  now
    let x be object;
    assume x in dom (f +* (i .--> X));
    per cases;
    suppose A4: x = i;
      then x in {i} by TARSKI:def 1;
      then x in dom({i} --> X) & x in dom({i} --> Y);
      then A5: x in dom(i .--> X) & x in dom(i .--> Y) by FUNCOP_1:def 9;
      then A6: (f +* (i .--> X)).x = (i .--> X).x by FUNCT_4:13
        .= X by A4, FUNCOP_1:72;
      (f +* (i .--> Y)).x = (i .--> Y).x by A5, FUNCT_4:13
        .= Y by A4, FUNCOP_1:72;
      hence (f +* (i .--> X)).x c= (f +* (i .--> Y)).x by A1, A6;
    end;
    suppose x <> i;
      then not x in dom(i .--> X) & not x in dom(i .--> Y) by TARSKI:def 1;
      then (f +* (i .--> X)).x = f.x & (f +* (i .--> Y)).x = f.x by FUNCT_4:11;
      hence (f +* (i .--> X)).x c= (f +* (i .--> Y)).x;
    end;
  end;
  hence thesis by A3, CARD_3:27;
end;
