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This was said at a lecture I attended:

$e$ is neutral element for operation $*$ if $\forall x (x*e=x \wedge e*x = x)$.

So, for example 0 is n. e. for disjunction and 1 is n. e. for conjuction (the truth value of disj./conj. equals that of $x$).

Conjuction ($\wedge$, neutral element $e = 1$) has inverse element because $\forall x(\neg x \wedge x = x \wedge \neg x = e)$. Notice the "$= e$" part. (Similiar thing was said for disjunction)

The problem is, 0 is $e$ (neutral element) for disjunction, not conjuction, and similarly for disjunction and 0.

My guess is that lecturer was talking about complements, not inverses, since there isn't always an element $y$ that could be found for any $x$ such that $x \wedge y = e = 1$ ($y$ would be an inverse since in conjunction with $x$ it would produce neutral element; there is no such y (for any $x = 0$)).

Am I missing something?

Thanks. :)

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There is almost always no $y$ such that $x\land y=1$. And usually there are many $y$ such that $x\land y=0$. From an algebraic point of view, the "right" addition is symmetric difference. –  André Nicolas Oct 6 '12 at 17:35
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My guess is that your lecturer was in fact talking about complements. For any $x$, there is a unique element $y$ such that both $x\wedge y=0$ and $x\vee y=1$--namely, $y=\neg x$. You're correct that for $x\neq 1$, there is no $y$ such that $x\wedge y=1$, and for $x\neq 0$, there is no $y$ such that $x\vee y=0$. However, when there is such a $y$--respectively, when $x=1$ and when $x=0$--we do have uniqueness--as we must respectively have $y=1$ and $y=0$.

It's also possible that your lecturer simply transposed conjunction and disjunction. In that case, for each $x$, there is a $y$ such that $x\vee y=1$, and for each $x$, there is a $y$ such that $x\wedge y=0$. Such $y$ aren't (generally) unique, though, so it's misleading to talk about inverses. We only get uniqueness if we require both $x\vee y=1$ and $x\wedge y=0$.

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