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

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

  • $\begingroup$ Wikipedia has a pretty good article about it. $\endgroup$ – MJD Apr 1 '14 at 13:13
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    $\begingroup$ The Wikipedia article seems a little imprecise and tentative. In particular, it states: A statement S is vacuously true if it resembles .... As far as I know, "resembles" does not usually have any precise meaning in math or logic or philosophy. Indeed I was inspired by exactly that Wikipedia sentence to post this question here. $\endgroup$ – Kenny LJ Apr 1 '14 at 20:52
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    $\begingroup$ @MJD: I don't think much of the WP article. It doesn't link its definitions to the sources it lists, and it does not relate to well-known related topics. The claim 'A statement S is "vacuously true" if it resembles the statement $P \Rightarrow Q$, where P is known to be false' is unattributed, imprecise (resembles?, no outermost quantifier?, does "known" mean this is an epistemic claim?) and I would say is false (rather, vacuous truths "resemble" a special case of this form), and the well-known example is generally described as a paradox of material implication. $\endgroup$ – Charles Stewart Apr 2 '14 at 13:43

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.

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    $\begingroup$ I see. The reason I ask is that when asked "Why is $\emptyset$ a subset of every set?", many answer that the statement "Every element of $\emptyset$ is an element of the set $A$" is vacuously true, therefore $\emptyset \subset A$. But given your answer to my present question, it would appear to me that giving such a "vacuous truth" explanation is no explanation at all---it seems to be just saying that $\emptyset \subset A$ because $\emptyset$ is the empty set. $\endgroup$ – Kenny LJ Mar 31 '14 at 21:45
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    $\begingroup$ @KennyLJ I see; what you there have is a not totally formal argument that $\emptyset \subseteq A$. If you want to formalize it (which in my opinion doesn't answer the question why?, but does show that you can nail down everything to the last dot), you have to resort to, for instance, axiomatic set theory. There you'd have an axiom such as $\exists e \forall x [ x \not \in e]$. Given such an $e$ and another $a$, you can then prove $\forall x [ x \in e \to x \in a ]$ (i.e. $e \subseteq a$). $\endgroup$ – Magdiragdag Apr 1 '14 at 5:17
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    $\begingroup$ @KennyLJ But the explanation 'Every element of ∅ is an element of the set A" is vacuously true' does make sense. You still have a statement of the form $\forall a \in \emptyset. P(a)$ which is true because we're quantifying over the empty set; that fact that $P(a)$ is the statement $a \in A$ doesn't matter. $\endgroup$ – Magdiragdag Apr 1 '14 at 7:18
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    $\begingroup$ @KennyLJ, to show that $\emptyset$ is a subset of $X$, assume $a \in \emptyset$ and prove $a \in X$. Since $a \in \emptyset$ is a contradiction, thus we may deduce anything. Hence $a \in X$. $\endgroup$ – goblin Apr 1 '14 at 15:32
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    $\begingroup$ One can understand $(\forall a\in X)(P(a))$ as being a shorthand for $(\forall a)(a\in X\to P(a))$. This view unifies the “quantification over the empty set” and “antecedent is always false” explanations of vacuous truth. $\endgroup$ – MJD Apr 1 '14 at 15:37

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.

  • $\begingroup$ Couldn't we also deduce that the empty set is not a subset of $A$ then? $\endgroup$ – Zduff Aug 27 '18 at 21:35
  • $\begingroup$ No. Given two sets $A$ and $B$, either $A$ is a subset of $B$ or it isn't. And until it is proved to be not a subset of $B$, it is one. If you want to prove that $\varnothing\nsubseteq A$, then you need to find a witness for that. But this witness is an element of the empty set which is not an element of $A$... $\endgroup$ – Asaf Karagila Aug 27 '18 at 22:32

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...

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    $\begingroup$ When I first prove to students that $\forall A(\varnothing\subseteq A)$ my experience is that first convincing them with a proof by contradiction is easy; then we can talk about vacuous truth and I can present the direct (or rather vacuous) argument. $\endgroup$ – Asaf Karagila Apr 1 '14 at 15:37
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    $\begingroup$ @DanielMak - NO : it is correct that $x \in \emptyset$ is always false, but the conditional $False \to P$ is true, for $P$ whatever. Thus $∀x(x \in \emptyset \to x \in B)$ is always true and this is exactly the reason why "the empty set is always subset of any set". $\endgroup$ – Mauro ALLEGRANZA Aug 1 '15 at 10:35
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    $\begingroup$ @H.R. - see Intuitionistic logic for rejection of "full" De Morgan's lasw and equivalence between $p \to q$ and $\lnot p \lor q$ as well as interdefinability of quantifiers. $\endgroup$ – Mauro ALLEGRANZA Aug 31 '16 at 18:53
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    $\begingroup$ @H.R. - I'm not saying that :-) In the context of my answer, I considered as example of "vacuous truth" the fact that $\emptyset$ is a subset of every set. It sounds "unnatural" and counterintuitive; so, why do not stay with a different "convention" ? The gist of my final cpomment in that it is not a convention at all: assuming the definition of $\emptyset$ and using he logical laws, we prove it. Thus, we cannot "deny" it, unless we change the definition or (in principle) we reject some of "usual" logical laws. 1/2 $\endgroup$ – Mauro ALLEGRANZA Sep 1 '16 at 7:08
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    $\begingroup$ Mauro I think it will be a little more clear to complete the transformation and say that $\lnot x\in B\equiv x\notin B$. $\endgroup$ – Masacroso Sep 12 '16 at 6:41

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.

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    $\begingroup$ I would suggest that you modify the claim that the opposite of "you're my favorite nephew" is "you're my least favorite nephew" $\endgroup$ – Dunham Feb 24 '17 at 16:18
  • $\begingroup$ Thanks @Dunham. I guess what you were getting at is that the actual "opposite" of that would be "you're not my favorite nephew", which would be false in this case? Anyway, whether that's what you meant or not, it would be a possible inaccuracy so I reworded. If it needs further clarification I welcome more feedback. $\endgroup$ – msouth Feb 25 '17 at 0:31
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    $\begingroup$ you're interpretation of my comment is correct. This is a nice intuitive explanation, I just thought it shouldn't stray too far from formal logic. $\endgroup$ – Dunham Feb 25 '17 at 15:26

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