Do LNC and LEM suffice to define negation?

February 17, 2015

What is negation?

One answer you find in the literature is that negation is the operator that makes each instance of the Law of the Excluded Middle (LEM) and the Law of Non-Contradiction (LNC) turn out to be true. That is, every sentence of the form $p\lor \neg p \qquad \neg(p\land\neg p)$ is true, no matter what sentence we use in the place of $$p$$ (where $$\neg$$ stands for negation, $$\lor$$ for disjunction, and $$\land$$ for conjunction).

This is the wrong way to try to define negation. If you read on, I’ll explain why.

Before I explain why, though, I should explain that taking LNC and LEM as defining negation isn’t an idiosyncratic view. If you read Laurence Horn’s entry on Contradiction in the Stanford Encyclopedia of Philosophy (and I recommend you do read it, it’s a great discussion!), you’ll see that he characterises negation in this way. Graham Priest’s chapter “Why Not? A Defence of a Dialethic Theory of Negation” takes this form of LNC and LEM to be central to defining what it is for one thing to be contradictory to another—in terms of which Graham characterises negation.

Here’s why it’s wrong to think that this form of LNC and LEM defines negation: It’s easy to see (once you check) that LNC and LEM don’t define anything. Altogether too many putative one-place connectives satisfy both LNC and LEM. Let’s suppose that $$\mathord\sim A$$ is just always interpreted as true, whatever $$A$$ happens to be. Then it turns out (if disjunction is interpreted in the usual way) that $$p\lor\mathord\sim p$$ is true (since its second disjunct is true) and so is $$\mathord\sim(p\land\mathord\sim p)$$ since it has the form $$\mathord\sim A$$. So, given this deviant “negations are always true” interpretation, LNC and LEM are always satisfied—by $$\sim$$, which isn’t negation.

This isn’t the only way one could interpret a sentential operator so as to satisfy LNC and LEM. For an even more deviant reading, let’s interpret $$\mathord\sim A$$ as $$\neg A\lor q$$ for any $$A$$ and for some sentence $$q$$. Then LNC and LEM for $$\sim$$-negation, that is, $p\lor (\neg p\lor q) \qquad \neg(p\land(\neg p \lor q))\lor q$ are themselves logically true (I leave the verification of this is an exercise for you). But $$\mathord\sim A$$—that is, $$\neg A\lor q$$—really isn’t the negation of $$A$$. (If we choose $$q$$ to be a tautology then $$\mathord\sim A$$ is simply equivalent to $$\top$$ which is the interpretation we saw before.)

In other words, if all we have to go on for interpreting “$$\sim$$” is that it satisfies LNC and LEM, then there many different ways it can be interpreted, including taking it to be highly contingent, where the semantic value of $$\mathord\sim A$$ depends on more than just the semantic value of $$A$$.

[EXCURSUS: If you’re prepared to go further afield, notice that in reflexive classical modal logics, we have both $$p\lor\Diamond\neg p$$ and $$\Diamond\neg(p\land\Diamond\neg p)$$ so satisfying LNC and LEM isn’t enough to tell us that we don’t mean “possibly not” by your so-called “negation”. And instead of $$\Diamond\neg$$ you could use $$\Diamond\Diamond\neg$$ or $$\Diamond\Diamond\Diamond\neg$$… etc.

To make a similar point in the context of first-order predicate logic, choose $$x$$ to be a variable at random, and read $$\mathord\sim A$$ as $$(\exists x)\neg A$$, whether $$x$$ is free in the formula $$A$$ or not. Notice that this very strange variable-binding “negation” still satisfies LNC and LEM, since $$A\lor (\exists x)\neg A$$ and $$(\exists x)\neg (A\land (\exists x)\neg A)$$ are tautologies, but $$(\exists x)\neg$$ is not negation. This bizarre creature binds a variable. Negation, as we traditionally understand it, doesn’t.]

So, if LNC and LEM in the traditional form don’t do enough by themselves to pin down negation, what can we say? Something I’ve argued elsewhere is that the right way to understand LNC and LEM is more like this: $\vdash p\lor \neg p \qquad p\land\neg p\vdash$ This LEM ($$\vdash p\lor \neg p$$) tells us that $$p\lor\neg p$$ is undeniable (to deny it is out of bounds), and this LNC ($$p\land\neg p\vdash$$) tells us that $$p\land\neg p$$ is unassertable (to assert it is out of bounds). This pair of rules does a much better job of pinning down the behaviour of negation.

Just how much better? Well, given a little more logic (the traditional sequent calculus rules for $$\land$$ and $$\lor$$, and the Cut rule), then if $$\neg$$ and $$\sim$$ both satisfy LNC and LEM in this form, then both negations are logically equivalent. Here’s a demonstration that $$\mathord\sim A\vdash\neg A$$ (you can generalise this to a proof of equivalence, though the dual proof of $$\neg A\vdash\mathord\sim A$$). Since $$A\land\mathord\sim A\vdash$$, by the usual behaviour of $$\land$$, we have $$A,\mathord\sim A\vdash$$. Since $$\vdash A\lor\neg A$$, by the usual behaviour of $$\lor$$, we have $$\vdash A,\neg A$$. From $$\vdash A,\neg A$$ and $$A,\mathord\sim A\vdash$$, by the Cut rule (cutting on $$A$$) we have $$\mathord\sim A\vdash\neg A$$.

So, if you are tempted by the idea that LNC and LEM together define negation, you should be careful to specify LNC as a denial or rejection rule ($$p\land\neg p\vdash$$, which says that $$p\land\neg p$$ is to be denied, or to be rejected), rather than $$\neg(p\land \neg p)$$ which isn’t strong enough to say what you want it to say—the two instances of negation gives you altogether too much wiggle room.

None of this is an argument that LNC and LEM actually hold. For that, you’ll need to look elsewhere.

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