Introduction to graph theory/Proof of Theorem 2

Statement
A graph is a forest if and only if for every pair of distinct vertices $$u,v$$, there is at most one $$u,v$$-path.

Proof
Suppose you have a graph $$G$$ which is not a forest. Since a forest is defined to be a graph containing no cycles, it is clear that $$G$$ must have a cycle $$C$$. All cycles have at least three vertices. Let $$v_1, v_2, \ldots, v_n$$ (with $$n\ge 3$$) be the vertices on $$C$$ in order. Then there are two distinct $$v_1,v_2$$-paths, namely $$v_1v_2$$ and $$v_1v_nv_{n-1}\ldots v_3v_2$$.

Now suppose that $$G$$ is a forest, but has two distinct $$u,v$$-paths, namely $$ua_1a_2\ldots a_nv$$ and $$ub_1b_2\ldots b_mv$$. We would like to say that these create a cycle $$ua_1\ldots a_nvb_mb_{m-1}\ldots b_1u$$. However, a cycle can only include each vertex once, and it is possible that one of the $$b_i$$ is equal to one of the $$a_j$$. To that end, write $$a_0=b_0=u$$ and $$a_{n+1}=b_{m+1}=v$$. Let $$i$$ be the largest value of $$i<=n$$ such that $$a_i=b_j$$ for some $$j$$. Then we find a cycle, namely $$a_ia_{i+1}\ldots a_nvb_mb_{m-1}\ldots b_j$$. By our choice of $$i$$, clearly no vertex appears more than once.