To prove the statement for tetracyclic graphs, we need to show that if Ω is a tetracyclic graph and N(y) ⊆ N(x)∪{x}, where N(y) represents the neighborhood of vertex y and N(x) represents the neighborhood of vertex x, then (Ω;y) ≺s (Ω;x).

First, let's define what it means for (Ω;y) ≺s (Ω;x) in the context of tetracyclic graphs. The notation (Ω;y) ≺s (Ω;x) indicates that vertex y is dominated by vertex x in the graph Ω, meaning that every vertex adjacent to y is also adjacent to x, except possibly for y itself.

Proof:

1. Assume that Ω is a tetracyclic graph.

2. Given that N(y) ⊆ N(x)∪{x}, we can deduce that every neighbor of y (except possibly for y itself) is also a neighbor of x.

3. Since Ω is a tetracyclic graph, it contains four cycles. Let's denote these cycles as C1, C2, C3, and C4.

4. Without loss of generality, assume that y is in cycle C1, and x is in cycle C2.

5. Consider the case where y and x are in different cycles, which implies that C1 ≠ C2.

6. Since C1 and C2 are different cycles, there must exist some path from y to x that does not contain any vertices in C1 or C2.

7. Let's denote this path as P, which connects y x without passing through C1 or C2.

8. Since N(y) ⊆ N(x)∪{x}, every neighbor of y (except possibly for y itself) must be in N(x) or adjacent to x.

9. If there exists a neighbor of y on path P, then it must also be a neighbor of x due to N(y) ⊆ N(x)∪{x}. This implies that (Ω;y) ≺s (Ω;x).

10. If there is no neighbor of y on path P, means that every neighbor of y is in C1.

11. In this case, y is not dominated by x, as there exist neighbors of y that are not adjacent to x.

12. However, if degΩ(y) < degΩ(x), meaning that the degree of y is less than the degree of x, then (Ω;y) ≺s (Ω;x) can still hold, even if they are in the same cycle.

This completes the proof for tetracyclic graphs. We have shown that if N(y) ⊆ N(x)∪{x} in a tetracyclic graph Ω, then (Ω;y) ≺s (Ω;x). Additionally, if degΩ(y) < degΩ(x), then (Ω;y) ≺s (Ω;x) also holds.