What Is Vacuous Truth?
Theorem
For any proposition $p$, a contradiction $c$, and $A_{\alpha} \subset X$, the following hold.
- [1] Vacuous truth: $c \implies p$
- [2] Union: $\displaystyle \bigcup_{\alpha \in \emptyset} A_{\alpha} = \emptyset$
- [3] Intersection: $\displaystyle \bigcap_{\alpha \in \emptyset} A_{\alpha} = X$
Explanation
For example, take the statement “God is dead.” What happens if God does not exist, meaning the premise itself is already wrong? If God does not exist, then $0$ gods have died, so it becomes true without even having to question who really died or lived. On the other hand, the statement “God is alive.” is also necessarily true if God does not exist, since we are checking $0$ individuals.
In this way, when the premise is a contradiction, the assertion becomes true no matter what it is, and this is called a vacuous Truth or a tautology. Of course, there are tautologies other than this form, but among them the hardest to accept is $c \implies p$. Similarly to the example above, given a set, we can take the union or intersection of $0$ of its subsets. Considering that $\displaystyle \sum_{n=0}^{0} n = 0$, it should become a little easier to accept.
Proof
Vacuous Truth
It suffices to show that $c \to p$ is true both when $p$ is true and when $p$ is false. Since $x \to y \equiv \lnot ( x \land \lnot y )$, $$ c \to p \equiv \lnot ( c \land \lnot p ) $$ whether $p$ is true or false, the result of taking the conjunction with $c$ is false, so $$ \lnot ( c \land \lnot p ) \equiv \lnot c $$ Since the negation of a contradiction $c$ is always true, $c \to p$ is true regardless of $p$.
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Union
It suffices to show that $\displaystyle x \notin \bigcup_{\alpha \in \emptyset} A_{\alpha}$ for all $x \in X$. $$ \begin{align*} x \notin \bigcup_{\alpha \in \emptyset} A_{\alpha} \iff & \lnot \left( x \in \bigcup_{\alpha \in \emptyset} A_{\alpha} \right) \\ \iff & \lnot ( x \in A_{\alpha_{0}} \text{ for some } \alpha_{0} \in \emptyset ) \\ \iff &x \notin A_{\alpha} \text{ for all } \alpha \in \emptyset \\ \iff & \alpha \in \emptyset \to x \notin A_{\alpha} \end{align*} $$ Since the set $\emptyset$ having an element contradicts the definition of the empty set, $\alpha \in \emptyset$ is false. By [1] vacuous truth, $\alpha \in \emptyset \to x \notin A_{\alpha}$ is true, and its equivalent $\displaystyle x \notin \bigcup_{\alpha \in \emptyset} A_{\alpha}$ is also true.
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Intersection
It suffices to show that $\displaystyle x \in \bigcap_{\alpha \in \emptyset} A_{\alpha}$ for all $x \in X$. $$ \begin{align*} x \in \bigcap_{\alpha \in \emptyset} A_{\alpha} \iff & x \in A_{\alpha} \text{ for all } \alpha \in \emptyset \\ \iff &\alpha \in \emptyset \to x \in A_{\alpha} \end{align*} $$ Again, since the set $\emptyset$ having an element contradicts the definition of the empty set, $\alpha \in \emptyset$ is false. By [1] vacuous truth, $\alpha \in \emptyset \to x \in A_{\alpha}$ is true, and its equivalent $\displaystyle x \in \bigcap_{\alpha \in \emptyset} A_{\alpha} $ is also true.
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