TITLE: Strong induction doesn't require a base case?
QUESTION [7 upvotes]: I'm considering the natural numbers to be the nonnegative integers.  The principle of strong induction can be stated as follows,
"If $P$ is a property such that for any $x$, if $P$ holds for all natural numbers less then $x$, then $P$ holds for $x$ as well, then $P$ holds for all natural numbers $x$."
This doesn't require a base case.  My question is why?  Shouldn't $P$ hold for $0$ for the induction to work?

REPLY [4 votes]: $P$ holds vacuously for every natural number smaller than $0$. Hence by the induction hypothesis, $P$ holds for $0$.

REPLY [3 votes]: If you know that $P$ holds for $x$ whenever $P$ holds for all $y$ with $y < x$, then the special case for $x = 0$ tells you that $P$ holds for $0$. This is because there are no $y < 0$, and so $P$ holds for all $y$ with $y < 0$ is vacuously true.