Is the inverse of MST cycle property always true? Why?
If you want to test whether a specific edge belongs to some MST, you can use the following property.

Claim. An edge $e$ belongs some MST if and only if for every $\epsilon &gt; 0$, if we reduce the weight of $e$ by $\epsilon$ then it belongs to all MSTs.

Proof. We first show that if $e$ belongs to some MST then for every $\epsilon &gt; 0$, if we reduce the weight of $e$ by $\epsilon$ then it belongs to all MSTs.

Suppose that MSTs in the original graph have weight $w$. When we decrease the weight of $e$ by $\epsilon$, only trees containing $e$ have decreased weight. Since $e$ belongs to some MST, the new graph has MSTs of weight $w - \epsilon$, and all of them contain $e$.

We complete the proof by showing that if $e$ doesn't belong to any MST then for small enough $\epsilon &gt; 0$, if we decrease the weight of $e$ by $\epsilon$ then $e$ still belongs to no MST. Indeed, suppose that MSTs in the original graph have weight $w$, and the lightest spanning tree containing $e$ has weight $w+\delta$ for $\delta &gt; 0$. If we decrease the weight of $e$ by less than $\delta$, it still belongs to no MST. $\qquad \square$

I'll let you figure out how to use the claim to test whether $e$ belongs to some MST; it suffices to find one MST in one graph. The same approach seems wasteful if you want to test all edges. I'll leave you to ponder whether any optimization is possible in that case.