theorem
  Free (p 'or' q) = Free p \/ Free q
proof
  thus Free (p 'or' q) = Free ('not' p '&' 'not' q) by Th60
    .= Free 'not' p \/ Free 'not' q by Th61
    .= Free p \/ Free 'not' q by Th60
    .= Free p \/ Free q by Th60;
end;
