Let $G(V, E)$ be a simple, undirected, edge-weighted graph with unique edge weights. Which of the following statements about the minimum spanning trees (MST) of $G$ is/are true?

Since all edge weights are distinct, graph $G$ has a unique Minimum Spanning Tree (MST).

• Option (A) is TRUE: By the Cycle Property of MSTs, for any cycle $C$ in a graph with distinct edge weights, the edge with the strictly largest weight in $C$ cannot belong to any MST.

• Option (B) is FALSE: The edge with the smallest weight in a specific cycle $C$ is not guaranteed to be in the MST. For example, consider a 3-cycle $A-B-C-A$ with edge weights $e(A, B) = 10$, $e(B, C) = 20$, and $e(A, C) = 30$. If there is another path between $A$ and $B$ via node $D$ with edge weights $e(A, D) = 1$ and $e(D, B) = 2$, then in the cycle $A-D-B-A$, the edge $e(A, B) = 10$ is the heaviest edge and will not be in the MST, even though it is the lightest edge in cycle $A-B-C-A$.

• Option (C) is FALSE: If a vertex $v$ has degree $1$ (a leaf node), its only incident edge is simultaneously the smallest and largest weight edge incident on $v$. This edge is a bridge and must be present in every spanning tree. More generally, the heaviest edge incident on a vertex can belong to an MST if it is required for connectivity (e.g., a bridge).

• Option (D) is TRUE: By the Cut Property of MSTs, consider the cut partition $(S, V \setminus S)$ where $S = \{v\}$. The edges crossing this cut are precisely all edges incident on vertex $v$. Since the edge with the smallest weight incident on $v$ is the unique minimum-weight crossing edge for this cut, it must be present in the MST.

Hence, the correct statements are A and D.