 For example if \(L\) is a first-order language with identity, whose signature consists of 1-ary predicate symbols, and \(L'\) is the language whose sentences consist of the four syllogistic forms (All \(A\) are \(B\), Some \(A\) are \(B\), No \(A\) are \(B\), Some \(A\) are not \(B)\) using the same predicate symbols, then \(L'\) is reducible to \(L\), because the syllogistic forms are expressible in first-order logic. (There are some quarrels about which is the right way to express them; see the entry on the traditional  square of opposition.)  But the first-order language \(L\) is certainly not reducible to the language \(L'\) of syllogisms, since in \(L\) we can write down a sentence saying that exactly three elements satisfy \(Px\), and there is no way of saying this using just the syllogistic forms. Or moving the other way, if we form a third language \(L''\) by adding to \(L\) the quantifier \(Qx\) with the meaning “There are uncountably many elements \(x\) such that …”, then trivially \(L\) is reducible to \(L''\), but the downward Loewenheim-Skolem theorem shows at once that \(L''\) is not reducible to \(L\).