Question about basis step of strong induction proof

Solution 1:

This type of induction works for any property $\,P(n)\,$ that is preserved under a shift, such as here where $\,P(n)\,$ true $\,\Rightarrow\,P(n\!+\!3)\,$ true. Writing $\,P\,$ for the subset of naturals where $\,P\,$ is true, the induction works as follows.

Theorem $\ $ Suppose $\,P\subseteq \Bbb N\,$ satisfies $\,n\in P\,\Rightarrow\, n\!+\!3\in P,\ $ for all $\,n\ge a.\ $ Then

$$\,a,a\!+\!1,a\!+\!\color{#c00}2\in P\,\Rightarrow\,n\in P{\rm\, \ for\ all\,\ } n\ge a$$

Proof $\ $ If not there is a least counterexample $\,\ell\not\in P.\,$ Note $\,\ell \ge a\!+\!\color{#c00}3\,$ so $\,\ell\!-\!3\ge a,\,$ Therefore, by our shift-closure hypothesis, $\,\ell = (\ell\! -\! 3)+ 3\in P,\,$ contradiction.

Remark $\ $ Clearly the proof generalizes from shift increment $\,k=3\,$ to arbitrary $\,k\ge 1,\,$ with $\,k\,$ consecutive integers $\,a,a\!+\!1,\ldots,a\!+\!k\!-\!1\,$ serving as the base cases, i.e. the foundation of the induction. Notice that the case $\,k=1\,$ is simply ordinary induction.