New to proofs and learning about strong induction. From what I understand strong induction allows us to say
$(S_1 \land S_2 \land S_3 \land \cdots \land S_k) \Rightarrow {{S_k}_+}_1$
However, in the book I'm reading (Book of Proof by Professor Richard Hammack) it seems the first example on page 162 contradicts the second example.
The first example deems it necessary in the basis case to ensure the equation holds for $S_1,S_2,S_3,S_4,S_5,S_6$. Which makes sense because the intent is to use ${{S_k}_-}_5$ in the proof.
Now, in the second example involving graphs strong induction is used yet again (page 164). The proposition is: If a tree has $n$ vertices, then it has $n-1$ edges. Now, the basis case only covers $n=1$ showing that a tree with $n=1$ vertex has $n-1=0$ edges.
But then in the inductive step and preceding remarks it is mentioned that: "... it was absolutely essential that we used strong induction in the above proof because the two trees $T_1$ and $T_2$ will not both have $k$ vertices. At least one will have fewer than $k$ vertices."
So my question is, how is that we did not need to show both $S_1$ and $S_2$ as a minimum in our basis case and then have our inductive hypothesis begin at $k\ge2$? If one of the trees has less than $k$ vertices and the starting point is arbitrary, could it not be that our random selection would be $k=1$ (i.e. $S_1$)? In which case $S_1 \Rightarrow S_2$ and now our two trees $T_1$ and $T_2$ must match up with $S_1$ and $S_0$ where $0$ is outside of $\mathbb{N}$?
