theorem
  Free(p 'or' q) = Free p \/ Free q
proof
  p 'or' q = 'not'('not' p '&' 'not' q) by QC_LANG2:def 3;
  hence Free(p 'or' q) = Free('not' p '&' 'not' q) by Th39
    .= Free 'not' p \/ Free 'not' q by Th42
    .= Free p \/ Free 'not' q by Th39
    .= Free p \/ Free q by Th39;
end;
