Theorem uvtxnbgr

Description: A universal vertex has all other vertices as neighbors. (Contributed by Alexander van der Vekens, 14-Oct-2017) (Revised by AV, 3-Nov-2020) (Revised by AV, 23-Mar-2021)

	Ref	Expression
Hypothesis	uvtxnbgr.v	$$⊢V=\mathrm{Vtx}\left(G\right)$$
Assertion	uvtxnbgr	$$⊢N\in \mathrm{UnivVtx}\left(G\right)\to G\mathrm{NeighbVtx}N=V\setminus \left\{N\right\}$$

Proof

Step	Hyp	Ref	Expression
1		uvtxnbgr.v	$$⊢V=\mathrm{Vtx}\left(G\right)$$
2	1	nbgrssovtx	$$⊢G\mathrm{NeighbVtx}N\subseteq V\setminus \left\{N\right\}$$
3	2	a1i	$$⊢N\in \mathrm{UnivVtx}\left(G\right)\to G\mathrm{NeighbVtx}N\subseteq V\setminus \left\{N\right\}$$
4	1	uvtxnbgrss	$$⊢N\in \mathrm{UnivVtx}\left(G\right)\to V\setminus \left\{N\right\}\subseteq G\mathrm{NeighbVtx}N$$
5	3 4	eqssd	$$⊢N\in \mathrm{UnivVtx}\left(G\right)\to G\mathrm{NeighbVtx}N=V\setminus \left\{N\right\}$$