I would like you to tell me if the proof is correct and how can I improve the formalisation of it. Also, if all the assumptions/steps of the proof are correct.
A tree has at least two leaves. (proof by contradiction.)
I intend to prove the above statement, and to do so I proceed by contradiction.
Let $T$ be a tree of order $n \geqslant 2$, in which we assume the number of leaves is $< 2$.
It must have at least one leaf, otherwise we would have a cycle. Then, let us call that leaf $l$.
Consider the path that starts at $l$, and follows to any other vertex $v$. If $v$ is a leaf, we are done. Otherwise it must be adjacent to at least another vertex $u$. Repeat the reasoning for $u$.
Thereby, our path can only end in a leaf (which is not possible by assumption) or in an already visited vertex creating a cycle (which is not possible by assumption since the graph we consider is a tree) and hence we found a contradiction.
P.S: sorry for the wrong math symbols,
A lot of thanks to all of you!
Best Answer
I'm convinced. Your proof looks good to me.
Another way to do this would be with a handshake argument. I think this argument is perhaps a bit easier to see. An equivalent definition of a tree with $n$ vertices is that it has $n-1$ edges. So let's assume exactly one leaf. Then every other vertex has degree at least $2$. So by the Handshake Lemma, $\sum_{v \in V(T)} d(v) = 2n - 2$. Since every vertex except the one leaf has degree at least $2$, we get $2 \sum_{v \in V(T)} d(v) \geq 2[2(n-1) + 1] = 4n - 2 > 2(n-1)$, a contradiction.