Proof of Correctness
Let P be a connected, weighted graph. At every iteration of Prim's algorithm, an edge must be found that connects a vertex in a subgraph to a vertex outside the subgraph. Since P is connected, there will always be a path to every vertex. The output Y of Prim's algorithm is a tree, because the edge and vertex added to tree Y are connected. Let Y1 be a minimum spanning tree of graph P. If Y1=Y then Y is a minimum spanning tree. Otherwise, let e be the first edge added during the construction of tree Y that is not in tree Y1, and V be the set of vertices connected by the edges added before edge e. Then one endpoint of edge e is in set V and the other is not. Since tree Y1 is a spanning tree of graph P, there is a path in tree Y1 joining the two endpoints. As one travels along the path, one must encounter an edge f joining a vertex in set V to one that is not in set V. Now, at the iteration when edge e was added to tree Y, edge f could also have been added and it would be added instead of edge e if its weight was less than e (we know we encountered the opportunity to take "f" before "e" because "f" is connected to V, and we visited every vertex of V before the vertex to which we connected "e" ). Since edge f was not added, we conclude that
Let tree Y2 be the graph obtained by removing edge f from and adding edge e to tree Y1. It is easy to show that tree Y2 is connected, has the same number of edges as tree Y1, and the total weights of its edges is not larger than that of tree Y1, therefore it is also a minimum spanning tree of graph P and it contains edge e and all the edges added before it during the construction of set V. Repeat the steps above and we will eventually obtain a minimum spanning tree of graph P that is identical to tree Y. This shows Y is a minimum spanning tree.
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