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 = ∅ → G NeighbVtx K = ∅

Proof

Step Hyp Ref Expression
1 nel02 ⊢ Vtx ⁡ G = ∅ → ¬ K ∈ Vtx ⁡ G
2 df-nel ⊢ K ∉ Vtx ⁡ G ↔ ¬ K ∈ Vtx ⁡ G
3 1 2 sylibr ⊢ Vtx ⁡ G = ∅ → K ∉ Vtx ⁡ G
4 eqid ⊢ Vtx ⁡ G = Vtx ⁡ G
5 4 nbgrnvtx0 ⊢ K ∉ Vtx ⁡ G → G NeighbVtx K = ∅
6 3 5 syl ⊢ Vtx ⁡ G = ∅ → G NeighbVtx K = ∅