Question about basis step of strong induction proof Let $P(n)$ be any collection of $n$ coins that can be obtained using a combination of $3$ cent and $5$ cent coins. Use strong mathematical induction to prove that $P(n)$ is true for all integers $n \ge 14$.


Basis: $P(14), P(15), P(16)$ can be made up of $5$ cents and $3$ cents. 

Inductive Step: $P(14)$ through $P(k)$ are true

$P(k + 1): k + 1 = (k + 1 – 3) + 3$ where if $k \ge 16$, then $(k + 1 – 3) \ge 14$ so $k + 1 \ge 17$, so $P(k + 1)$ is true by hypothesis. 

If this makes sense, why do we need to show the basis steps explicitly? Aren't we assuming them to be true in the inductive step?
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.
Yes, you are assuming them to be true in the later step. (But you can assume anything you want.) 

Precisely to make the argument valid you need to establish (or to prove) the base cases somewhere. 

How do you know that $P(17)$ is true? You know it because you can reduce it to the truth of $P(14)$ (this is what is done in the induction step), and the truth of this was established explicitly. 

If you do not do this, then you can only say $P(17)$ is true if $P(14)$ is true. (This is the induction step.) But you need to show the latter somewhere. 

Of course, and this is the charm of induction, you can do longer chains. 

Why is $P(20)$ true? It is true if $P(17)$ is true. And $P(17)$ is true if $P(14)$ is true. And $P(14)$ is in fact true. So $P(20)$ is true.
Yet, if you do not know the truth of $P(14)$  then you cannot conclude anything. 

To look at it differently. You could just as well do the argument for the induction step with $k \ge 3$ and assuming $P(1), P(2), P(3)$ true. But you cannot get $4$ with $3$ and $5$ coins! Why? because you cannot get $1$ either.  But under the (false) assumption you can get $1$, you could also get $4$. 

So you need to check that your base assumption are not false. This is what the base step is for.