# why can only universal statements be vacuously true?

So I'm new to logic and im getting really confused on definitions namely that of universal and existential quantifiers and why vacuous truths only apply to statements with a universal quantifier $$(\forall)$$ and not for existential quantifier $$(\exists)$$.

I read that the 2 statements

(1) $$\forall x\in A\,P(x)$$
(2) $$\forall x(x\in A\implies P(x))$$
Are logically equivalent but why cant we say the same thing about the existential quantifier namely why cant we say that statements $$(1)$$,$$(2)$$ can be used.

(1) $$\exists x\in A\,P(x)$$
(2) $$\exists x(x\in A\implies P(x))$$

So why do we say that only universal or conditional statements are vacuously true?.

The formula $$\exists x \in A\;P(x)$$ stands for $$\exists x(x \in A \land P(x))$$. This is because we want the following chain of equivalences to hold:

\begin{align*} \exists x \in A\;P(x) &\equiv \lnot \forall x \in A\;\lnot P(x) \\ &\equiv \lnot \forall x(x \in A \implies \lnot P(x)) \\ &\equiv \exists x(x \in A \land P(x)) \end{align*} where the first equivalence is the desired De Morgan law, the second follows from the definition as you give it and the third follows because to falsify $$\phi \implies \psi$$ you have to make $$\phi$$ true and $$\psi$$ false.

$$\exists x(x \in A \implies P(x))$$ would be trivially true (unless $$A$$ is the entire universe of discourse) since for $$x \not\in A$$, $$x \in A \implies \phi$$ is true for any $$\phi$$.

The analogue of the vacuously true universal: $$\forall x \in \emptyset\;P(x)$$ is the vacuously false existential: $$\exists x \in \emptyset\;P(x)$$.

• why is it for A being the entire universe of discourse?. Commented Feb 14, 2021 at 15:22
• If $A$ is the entire universe of discourse, then $x \in A$ is true for any $x$, and so $x \in A \implies P(x)$ is equivalent to $P(x)$. Commented Feb 14, 2021 at 15:29
• I don't understand what you mean by "the same or construction". Commented Feb 14, 2021 at 20:40
• Well you could, but it will be guaranteed to be false unless $A$ is the universe of discourse (compare how using $\implies$ rather than $\land$ in the definition of $\exists x \in A \; P(x)$ gives something which is trivially true unless $A$ is the universe of discourse, as discussed in my answer). Try seeing where you get starting with the De Morgan law $\forall x \in A\; P(x) \equiv \lnot \exists x\in A \; \lnot P(x)$ using the definition we have now agreed for $\exists x \in A\;\lnot P(x)$. Commented Feb 14, 2021 at 23:05
• Because there is no point in having notations like $\forall x \in A\;P(x)$ or $\exists x \in A\;P(x)$ if the constraint $x \in A$ means that the formula will always be true or always be false. Try writing down in English what you want these formulas to mean intuitively (every element of $A$ satisfies $P$; some element of $A$ satisfies $P$): the definitions we have arrived that match those intuitions. Commented Feb 14, 2021 at 23:17

As I understand it, the implication $$A\to B$$ is said to be vacuously true if $$A$$ is false. We can see this from lines 3 and 4 of the truth table for implication:

$$\neg A \to (A\to B)$$ is also a tautology:

This can also be proven using a form of natural deduction:

Vacuous truth doesn't necessarily have anything to do with quantifiers or empty sets, though it is used, for example, to show that the empty set is unique, i.e. that all empty sets are equal. It is also used in proofs by cases when one or more of the cases being considered are proven or assumed to be false.