I want to prove that some function $f(x)$ assigns the same quantity $y$ to any pair of identical objects in a finite set of $|N|$ objects. Given the structure of the problem, I'm using induction. So far, I'm proving that my claim is true under the $2$ objects case; and I then do the inductive step. However, for the case $|N|=1$, the statement is vacuously true. Therefore, my original statement is vacuously true for the case $|N|=1$.
Then, my question is the following one: can I use the case of $|N|=1$, which holds vacuously, as the first step of my inductive proof?
I hope my question makes (now) sense.
Thank you all very much in advanced for your time.
The short answer is yes -- vacuous truth is truth. It's no different from any other truth.
However, you have to be careful to make sure your inductive step isn't implicitly relying on anything which doesn't apply in the non-vacuous case.
For example, suppose I wanted to prove, for a finite set of $N$ colored objects, that every pair of objects is blue. I prove this as follows: for $N=1$, this is vacuously true. Now suppose among every set of $N$ objects, every pair is blue. Among a set of $N+1$ objects, each pair of objects belongs to a subset of $N$ objects, so they must be blue; therefore every pair of objects in the set of $N+1$ is blue. QED.
Clearly this argument can't be sound. The problem is the claim "Among a set of $N+1$ objects, each pair of objects belongs to a subset of $N$ objects." This isn't true when $N=1$ and $N+1=2$. The (only) pair of objects in the set of $2$ doesn't belong to a subset of size $1$. The claim will hold only for $N\geq2$.
This isn't a problem with using vacuous truth per se, but it's an example of the type of thing that could go wrong when you try to jump from the vacuous case to the non-vacuous case. If you make sure to avoid this sort of fallacy, inductive arguments with a vacuous base case are perfectly okay.