What precisely is a vacuous truth?

Question

Is there a proper and precise definition that goes something like this?

Definition. A statement $S$ is a vacuous truth if ... ...

Answer 1

No. The phrase "vacuously true" is used informally for statements of the form $\forall a \in X: P(a)$ that happen to be true because $X$ is empty, or even for statements of the form $\forall a \in X: Q(a) \to P(a)$ that happen to be true because no $a \in X$ satisfies $Q(a)$. In both cases, it is irrelevant what statement $P(a)$ is.

I guess you could turn this into a formal definition of a property of statement, but that's not standard.

Answer 2

We say that an implication $p\to q$ holds vacuously if $p$ is always false. That is to say, it is impossible to have $p$ true and $q$ false. So the implication is a tautology.

Of course tautologies exist in propositional calculus, and not quite in predicate logic (and thus not in first-order logic), but the concept caries over.

So when we say that the empty set is a subset of $A$ is vacuously true, we say that there is just no counterexample to the contrary. Why is that true? Because the set is empty.

Answer 3

You are "not alone" with your doubt about $\emptyset$; see the "debate" in this post.

You must "work with" Asaf's answer: basically, we have the definition of $\emptyset$ and that of inclusion :

$A \subseteq B =_{def} \forall x (x \in A \rightarrow x \in B)$.

We have also a "basic principle" of mathematical reasoning (but not only) : "stay with the consequences of your assumptions, also when they are (a little bit) counterintuitive, unless you have found a contradiction (or a more satisfying theory)".

Let us try the "exercise" of negate the definition of set-inclusion : from $\lnot \forall x (x \in A \rightarrow x \in B)$, due to the fact that $\lnot \forall$ is equivalent to $\exists \lnot$ and that $p \rightarrow q$ is equivalent to $\lnot p \lor q$, we may "translate" the above formula into : $\exists x \lnot (\lnot x \in A \lor x \in B)$.

The final passage is with De Morgan, i.e.: $\lnot (p \lor q) \equiv (\lnot p \land \lnot q)$ and double negation, i.e.$\lnot \lnot p \equiv p$. Thus, we may transform the above formula into $\exists x (x \in A \land \lnot (x \in B))$.

Now we apply it with $\emptyset$ in place of $A$ :

$\exists x (x \in \emptyset \land x \notin B)$.

What does it means ? That there exists an object $x$ that belongs to $\emptyset$ and ...

But we have no elements into $\emptyset$; thus, the "purported" negation of $\emptyset \subseteq A$ must be always false.

This is the "reason why" $\emptyset$ is a subset of every set.

In the above argument we have used the rules of logic: some of them are "refused" by some (few) mathematicians. You may not accept some (all) of them : it's up to you; in this way you may try to "escape" from asserting the unwanted property of the emptyset...

Answer 4

Not sure if this is as formal as you want, but it might be a step toward a formalization.

I generally think of it like this:

A statement S is a vacuous truth if ...

S is formally true, but does not convey any information.

The example I like for the layman is the "You're my favorite nephew" joke that is said by a person with only one nephew. People understand it--what makes it funny is that it's true, but it doesn't mean anything. (For example, this statement would also be true--"You're my least favorite nephew".)

To formalize it a bit, you can define favorite as follows: "List all your nephews in order of preference, highest preference first. The first nephew on the list is your favorite." Then, in a list of one, that one is the favorite. Similarly, "List all your nephews in order of preference, highest preference first. The last nephew on the list is your least favorite."

The aunt is not lying--she is perfectly correct, by definition, in stating that the nephew is her favorite. However, we would say that her statement "doesn't tell us anything". In this case, it doesn't tell us anything because she has no other nephews. What the nephew wants to think is "there is someone out there that she considers me superior to", but what turns out to be the case is that the nephew's reasoning is about an empty set--the set of nephews that she likes less than him.

From there, you might be able to get someone to see the (vacuous) truth of "Every person that has walked on the surface of the sun has survived." Formally, it's accurate. It's also formally accurate to say "Every person that has walked on the surface of the sun died instantly." While both statements are formally true, they convey no information.

It might be too restrictive to say S only tells us about members of the empty set--because S might not tell us "things", or maybe not things "about members" of any set--but that would, I think, describe a class or collection of vacuously true statements.

Answer 5

An implication is said to be vacuously true if its antecedent is false.

$$\neg A \implies (A\implies B)$$

This logical principle is a tautology, and can be used to introduce an arbitrary consequent in an implication.

Example 1: If pigs can fly, then the President is a genius.

Proof: The antecedent, "pigs can fly," is false. Therefore the above implication is true, whether or not the consequent, "the President is a genius," is true. Note: We cannot infer from this that the President is not a genius. Nor can we infer the contrary that he is a genius.

Example 2: $\forall x: (x\in \emptyset \implies 0=1)$

Proof: $y\in \emptyset$ is always false. Then the implication $y\in \emptyset \implies 0=1 $ is true (vacuously so). Generalizing, we obtain as required: $$\forall x: (x\in \emptyset \implies 0=1)$$

Answer 6

Contrary to some of the existing answers, I don't have the impression that one typically speaks of a vacuous truth if the statement is a pure implication whose antecedent happens to be false; the usual use of "vacuous truth" occurs in the context of universal claims where the antecedent is always (i.e. for every object) false.

A universally quantified implication is vacuously true iff it is true just because there are no objects to satisfy the antecedent. Formally,

A statement of the form $\forall u (\phi \to \psi)$ (where $u$ is a varible and $\phi$ and $\psi$ are arbitrary formulas) is vacuously true in a structure $\mathcal{S} = \langle \mathcal{D}, \mathcal{I} \rangle$ iff
for all $a \in \mathcal{D}$ and arbitarary $v$: $\mathcal{S}, v^{[u \mapsto a]} \not \models \phi$.