theorem Th101:
  for x, y, z st x => y is LD-provable & y => z is LD-provable holds
      x => z is LD-provable
proof
  let x, y, z;
  assume that
    A1: x => y is LD-provable and
    A2: y => z is LD-provable;
  x = x '&' y by A1, Th92
      .= x '&' (y '&' z) by A2, Th92
      .= (x '&' y) '&' z by Th98
      .= x '&' z by A1, Th92;
  hence thesis by Th92;
end;
