# Prove that a tree with two vertices of degrees $k$ and $l$ has at least $k + l - 2$ leaves

Let there be a tree with at least two vertices. One vertex has degree $$k$$ and the other has degree $$l$$. Prove that such tree has at least $$k + l - 2$$ leaves.

My logic is that from one vertex you can reach at least $$k$$ leaves and from the other vertex you can reach at least $$l$$ leaves. But how exactly would I prove that you have to subtract $$2$$? Is it because if you start from one of the given vertices and go through the second one, then one of the leaves will be a duplicate?

• If you start from one of the vertices and follow a path of vertices to the second, you have discovered a cycle. And cycles are not allowed in trees. – Matthew Leingang May 13 at 18:00

In the tree $$T$$ there's a path from the vertex $$v$$ of degree $$k$$ to the vertex $$w$$ of degree $$l$$. Let this path start with the edge $$e$$. There are $$k-1$$ other edges from $$v$$, and one can walk from $$v$$ starting at any of these edges and reach a leaf. This accounts for $$k-1$$ leaves. Likewise start at $$w$$ to reach $$l-1$$ further leaves.
• A relevant (I think) addition is that the path from $v$ to $w$ is unique in a tree. So OP's concern about following a branch along an edge other than $e$ from $v$ and accidentally ending up at $w$ won't happen. – Matthew Leingang May 13 at 18:04