Consider a simple undirected weighted graph $G$, all of whose edge weights are distinct. Which of the following statements about the minimum spanning trees of $G$ is/are TRUE?

Since all edge weights in $G$ are distinct, $G$ has a unique Minimum Spanning Tree (MST). Let the edges in increasing order of weight be $e_1, e_2, e_3, e_4, \dots$.

• (A) is TRUE: By Kruskal's algorithm, $e_1$ is always chosen. Since $G$ is a simple graph, $e_2$ cannot form a cycle with $e_1$ (a cycle requires at least 3 edges in a simple graph). Thus, the edge with the second smallest weight ($e_2$) is always included in the MST.

• (B) is TRUE: $e_3$ can only create a cycle if $e_1, e_2, e_3$ form a triangle ($K_3$) on 3 vertices. In this case, $e_3$ is discarded. However, since the subgraph spanned by $\{e_1, e_2\}$ has only 3 vertices, any 4th distinct edge $e_4$ must be incident to at least one new vertex or cannot form a cycle solely with $\{e_1, e_2\}$ (since a 3-vertex simple graph can have at most 3 edges). Thus, if $e_3$ is rejected, $e_4$ is guaranteed to be accepted. Therefore, one or both of $e_3$ and $e_4$ must belong to the MST.

• (C) is TRUE: This is the standard Cut Property of MSTs: for any non-trivial cut $(S, V \setminus S)$, the edge of minimum weight crossing the cut must belong to every MST (and here, the unique MST).

• (D) is FALSE: When all edge weights in a connected undirected graph are distinct, the MST is always unique, so $G$ cannot have multiple minimum spanning trees.

Hence, the correct options are A, B, and C.