theorem Th105:
  for x, y, z holds x 'or' (y '&' z) = ((x 'or' y) '&' (x 'or' z))
proof
  let x, y, z;
  thus x 'or' (y '&' z) = 'not' (('not' x) '&' (('not' y) 'or' ('not' z)))
      .= 'not' ((('not' x) '&' ('not' y)) 'or' (('not' x) '&' ('not' z)))
          by Th104
      .= ((x 'or' y) '&' (x 'or' z));
end;
