Theorem:
Let be $G$ a weighted graph in which every edge has a different weight.
Suppose that $G$ has two spanning trees $A$ and $A'$. Let be $i$ the first index such that $e_i\ne e'_i$ and $i\ge 3$.
Suppose that $w(e_i)<w(e'_i)$ and consider that $A' \cup \{e_i\}$ has a unique cycle.
For each edge of this cycle then $w(e')<w(e_i)$ (else we would be able to create a tree $(A' \cup \{e_i\})\setminus\{e'\}$ of lower weight than $A'$, which contradicts the fact that $A'$ has minimal weight.
Therefore, every edge of this cycle must be in $A'$. Which contradicts the fact that A is a tree
I don't understand the conclusion in italics. This a demonstration from my teacher but it seems that it pop out without any explanation... Can you help me?
e'($e'$) rather thane\prime($e\prime$). This works with subscripts, too:e'_iande_i'both give $e'i$, as distinct from `e{i'}`, which gives $e_{i'}$. – David Richerby Nov 30 '15 at 08:43