Final answer:
The statement is false;
increasing edge capacities by one does not guarantee the minimum (s, t)-cut of the original graph will remain minimal in the new graph.
A counterexample with a simple directed graph demonstrates that a previously non-minimal cut can become minimal after increasing capacities.
Step-by-step explanation:
The question posits that if you have a graph G with a minimum cut S, and you create a new graph G' with the same structure but with increased capacities for all edges by one, then S will also be a minimum cut in G'.
The truthfulness of this statement can be determined by looking at how the minimum cut is defined and how increasing edge capacities may affect it.
By definition, a minimum cut in a graph is a partition of the vertices into two disjoint subsets that separates the source s from the sink t such that the sum of the capacities of the edges crossing the cut is minimized.
Upon increasing each edge's capacity by one, it does not necessarily mean that the original minimum cut will remain the minimum; especially if there is an alternative cut with more edges which were close to being a minimum cut in the unmodified graph.
Increasing all edge capacities by one could decrease the cut-capacity difference between the original minimum cut and other cuts.
Here is a counterexample: Consider a directed graph with vertices s, a, b, t and edges (s, a), (a, b), (b, t), and (s, t) with capacities as follows: c(s, a) = 2, c(a, b) = 2, c(b, t) = 2, and c(s, t) = 1.
The minimum s-t cut here is the edges (s, a) and (b, t) with total capacity 4. Now increase the capacity of each edge by one. The previously non-minimum cut consisting of just the edge (s, t) now has a capacity of 2, and it becomes a minimum cut because the total capacity across the cut (s, a) and (b, t) became 6, which is higher than 2. Thus, S is not a minimum s-t cut in G'.