theorem
  for x, y holds (x => y is LD-provable & y => x is LD-provable) iff x = y
proof
  let x, y;
  thus (x => y is LD-provable & y => x is LD-provable) implies x = y
  proof
    assume that A1: x => y is LD-provable and A2: y => x is LD-provable;
    thus x = x '&' y by A1, Th92 .= y by A2, Th92;
  end;
  assume x = y;
  hence x => y is LD-provable & y => x is LD-provable by Th92;
end;
