Why is this considered a proof by strong induction? I am currently working my way through The Book of Proof and I fail to fully grasp the concept of strong induction.
Normal induction makes perfect sense to me, I prove the base case, then I can assume $ \mathcal P(n) $ to be true and prove that $ \mathcal P(n)\to\mathcal P(n+1) $ is true, hence the proof is complete as I can just plug in the first value and the dominoes start falling so to speak.
To my understanding strong induction should work in the same fashion except that we have to proof not only the base case, but a sufficient number of cases and only then can we continue. (For example if I prove that a statement is true for the first 4 cases I can then finish my proof by showing that $ \mathcal P(n-3)\to\mathcal P(n+1) $ is true.) However, in the following proof it feels like this step is being skipped?
Suppose $ A_1,A_2,...A_n $ are sets in some universal set $ U $, and $ n≥2 $. Prove that $ \overline{A_1∩A_2∩···∩A_n}=\overline{A_1}∪\overline{A_2}∪···∪\overline{A_n}. $
Proof.

*

*The proof is by strong induction. When $ n=2 $ the statement is $
    \overline{A_1∩A_2}=\overline{A_1}∪\overline{A_2} $.  This is not an
entirely obvious statement, so we have to prove it [...] I am skipping this part here.

*Let $ k≥2 $. Assume the statement is true if it involves $ k $ or
fewer sets. Then

$$ \overline{A_1∩A_2∩···∩A_{k−1}∩A_k∩A_{k+1}}= $$ $$ \overline{A_1∩A_2∩···∩A_{k−1}∩(A_k∩A_{k+1})}= \overline{A_1}∪\overline{A_2}∪···∪\overline{A_{k−1}}∪\overline{A_k∩A_{k+1}}= $$ $$  \overline{A_1}∪\overline{A_2}∪···∪\overline{A_{k−1}}∪\overline{A_k}∪\overline{A_{k+1}}.$$
Thus the statement is true when it involves $ k+1 $ sets. This completes the proof by strong induction.
I can see that $ \overline{A_k∩A_{k+1}} = \overline{A_k}∪\overline{A_{k+1}} $ as we just proved this in the step above, but why can we assume that $ \overline{A_1∩A_2∩···∩A_{k−1}} = \overline{A_1}∪\overline{A_2}∪···∪\overline{A_{k−1}} $ for values $k>2$ ?
 A: We can try to analyse strong induction the same way as normal induction (in fact they are equivalent in "strength"), "step by step".
How does the above prove, for example, the case $n = 5$?
We have the base case $n = 2$.
For $n = 3$, we only use $n = 2$, and the result is proved in the induction step.
Now for $n = 4$, we have the results for $n = 2, 3$, and the induction step uses both and proves it.
If now we analyse $n = 5$, we have already seen that the result is true for $n = 2,3,4$, so the induction step can use all these previous results and prove that case.
Ad infinitum.
A: The proof of the $n=k+1$ case uses the $n=k$ case (the first $k-1$ of its arguments being $A_i$ with $1\le i\le k-1$; the $k$th is $A_k\cap A_{k+1}$) followed by the $n=2$ case (with arguments $A_k,\,A_{k+1}$). Hammack seems to consider any proof whose inductive step uses more than the case $n=k$ to be strong.
A: With the comments I made under the OP's question in mind, here is a more precise way to present their logic (sketch out the regular base/step induction):
Step Case: Assume true for $k$ sets.
Let $B = A_k \cap A_{k+1}$
Then
$$\tag 1 A_1 \cap A_2 \cap \dots \cap A_{k-1} \cap A_{k} \cap A_{k+1} =  A_1 \cap A_2 \cap \dots \cap A_{k-1} \cap B$$
The $\text{rhs}$ is the intersection of $k$ sets so by our inductive hypothesis,
$$\tag 2 \overline{ A_1 \cap A_2 \cap \dots \cap A_{k-1} \cap B}  = \overline A_1 \cup \overline A_2 \cup \dots \cup \overline A_{k-1} \cup \overline B= $$
Since we've shown that $ \overline B = \overline A_k \cup \overline A_{k+1}$ (the base case $n = 2$),
substituting $\overline A_k \cup \overline A_{k+1}$ into $\text{(2)}$ proves the inductive step.
