I was pondering a bit more about this question regarding being able to "omit" the base case in a proof by strong induction due to vacuous truth. The post states:

Strong induction proves a sequence of statements $P(0), P(1), …$ by proving the implication

"If $P(m)$ is true for all nonnegative integers $m$ less than $n$, then $P(n)$ is true."

for every nonnegative integer $n$. There is no need for a separate base case, because the $n=0$ instance of the implication is the base case, vacuously.

However, if we consider $n=0$, we would have that the statement is vacuously true, which I would take to mean that the implication is true *regardless* of the validity of $P(0)$. However, clearly it's necessary for $P(0)$ to hold for an induction proof to be valid. So I'm confused on how, by omitting the base case, $n=0$ isn't just a tautology, making the implication true regardless of whether $P(0)$ actually holds.