For $G_1$, $G_2$, $G_3$ simple undirected graphs on the same vertices with disjoint edge sets, if $G_1\cup(G_2\cap G_3)=G_2$, then $G_1=G_2$

Question

If $G_1, G_2$ and $G_3$ are simple undirected graphs on the same set of vertices with disjoint edge sets. If we have a graph equation $$G_1\cup(G_2\cap G_3)=G_2$$ Then we have to show that $G_1=G_2,$ where $\cup$ and $\cap$ are graph union and intersection, respectively.

Note that these graph operations satisfy Boolean algebraic properties.

How to prove $G_1=G_2?$

Answer 1

The only way this can be satisfied with the graphs having disjoint edge sets is if $G_1$ and $G_2$ are both empty graphs (and hence equal, since they have the same vertex set).

This is because any edge of $G_1$ is an edge of $G_1\cup(G_2\cap G_3)$, but not of $G_2$, contradicting the relationship. So $G_1$ has no edges, and $G_1\cup(G_2\cap G_3)=G_2\cap G_3$. Since $G_2$ and $G_3$ have disjoint edge sets their intersection has no edges, and so if the relationship holds $G_2$ has no edges.

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Without the condition that the graphs have disjoint edge sets, it wouldn't be true. For example, let the graphs have vertex set $\{x,y,z\}$ with $E(G_1)=\{xy\}, E(G_2)=\{xy,yz\}$ and $E(G_3)=\{yz\}$.

Answer 2

Defining $G_0$ as the graph with the common vertex set but no edges, we can write $$G_2 = G_1\cup (G_2\cap G_3) = G_1\cup G_0 = G_1$$ Done!

Note: We can go on to observe that the edge sets of $G_1$ and $G_2$ must also be empty, but this isn't required to establish the equality.