Meaning of Strong Induction Hypothesis Variable I thought strong induction relied on a proposition being true for all (natural) numbers from a given one up to (say) k, then showing p(k+1) must be true.
For example, I’m trying to understand this proof: https://math.stackexchange.com/a/2096485/442515
In the inductive hypothesis:
Assume $m\in\mathbb{Z}^+$ with $1\le m \le k$ and that our proposition holds for $m$.
If the role of the $m$ there is not to establish that the proposition is true from 1 up to k, then what role does it actually play please? The hypothesis does not seem equivalent to saying "assume the proposition holds for all positive integers from 1 to k inclusive."
At this stage, observations that strong and weak induction are the same are unlikely to help. I feel I need to get my head around this "p(all values from 1 to k inclusive) implies p(k+1) idea first, as it's the only thing I've read about string induction that actually makes sense to me at my current level of understanding.
 A: Say that the statement to be proved is $S$, a statement about $n$, and that $S(m)$ means that the statement is true for $n=m$.
As an example from the proof you have linked to: In the proof the assumption $S(m)$ for $m: 1 \leq m \leq k$ is used to draw the conclusion that the statement, in particular, holds for $\dfrac{k+1}{2}$, that is, $S\Big(\dfrac{k+1}{2} \Big)$. This  in turn is used to prove $S(k+1)$.
So while the proof only uses the assumption $S\Big(\dfrac{k+1}{2} \Big)$ in order to prove $S(k+1)$, strong induction provides a straight forward way to let us assume  $S\Big(\dfrac{k+1}{2} \Big)$ in the first place, and in particular, weak induction would not have allowed us to do this.
I imagine it would be difficult to come up with some kind of induction proof that as its inductive hypothesis only assumes $S\Big(\dfrac{k+1}{2} \Big)$ in order to prove $S(k+1)$. Even if it was possible, why should one do this when strong induction provides a straight forward way for us to assume $S\Big(\dfrac{k+1}{2} \Big)$, and does not pose any other difficulties?
Edit:
After reading your comment and re-reading the linked proof, I am inclined to agree with you. I didn't read your question carefully enough the first time, sorry about that.
I just think that it is a bit of a sloppy formulation in the linked proof. What the author should mean, is indeed this:
"Assume that our proposition is true for all $m$ such that $m \in \mathbb{Z}_+, 1 \leq m \leq k$.
It is probably just a case of being used to how proofs by induction are usually formulated, so I guess most people (including myself) assume that the author really talks about all $m$ such that... not a single $m$ with those properties. And this really should be what the author means, because the proof does not work otherwise.
