I was taught that vacuously true statements are of the form $P \implies Q$ where $P$ is a false statement. This property is routinely used to make claims about the empty-set.
Consider the function $f: \emptyset \to A$ where $A$ is a non-empty set. We have that $f$ is a function. Consequently, it satisfies the condition that $(\forall x \in \emptyset)(\exists y \in A) (x,y) \in f$. It is contended that this condition is satisfied vacuously.
How would I parse the above condition in an if-then statement so as to be able to apply the properties of material implication and conclude the vacuousness of the statement?
Let $A,B,f$ be sets.
Then $f$ is a function $A\to B$ if and only $f\subseteq A\times B$ and $\forall x\left[x\in A\to\exists!y\left[\left(x,y\right)\in f\right]\right]$
If $A=\varnothing$ then $f=\varnothing$ acccording to the first condition.
Further we observe that: $$x\in\varnothing\to\exists!y\left[\left(x,y\right)\in f\right]$$ is vacuously true, and this for every $x$.