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Metamath Proof Explorer

Theorem nbgr0vtx

Description: In a null graph (with no vertices), all neighborhoods are empty. (Contributed by AV, 15-Nov-2020) (Proof shortened by AV, 10-May-2025)

		Ref	Expression
	Assertion	nbgr0vtx	$⊢ Vtx (G) = \emptyset \to G NeighbVtx K = \emptyset$

Proof

Step	Hyp	Ref	Expression
1		nel02	$⊢ Vtx (G) = \emptyset \to \neg K \in Vtx (G)$
2		df-nel	$⊢ K \notin Vtx (G) \leftrightarrow \neg K \in Vtx (G)$
3	1 2	sylibr	$⊢ Vtx (G) = \emptyset \to K \notin Vtx (G)$
4		eqid	$⊢ Vtx (G) = Vtx (G)$
5	4	nbgrnvtx0	$⊢ K \notin Vtx (G) \to G NeighbVtx K = \emptyset$
6	3 5	syl	$⊢ Vtx (G) = \emptyset \to G NeighbVtx K = \emptyset$