Metamath Proof Explorer


Theorem nbgrel

Description: Characterization of a neighbor N of a vertex X in a graph G . (Contributed by Alexander van der Vekens and Mario Carneiro, 9-Oct-2017) (Revised by AV, 26-Oct-2020) (Revised by AV, 12-Feb-2022)

Ref Expression
Hypotheses nbgrel.v ⊢ V = Vtx ⁡ G
nbgrel.e ⊢ E = Edg ⁡ G
Assertion nbgrel ⊢ N ∈ G NeighbVtx X ↔ N ∈ V ∧ X ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e

Proof

Step Hyp Ref Expression
1 nbgrel.v ⊢ V = Vtx ⁡ G
2 nbgrel.e ⊢ E = Edg ⁡ G
3 1 nbgrcl ⊢ N ∈ G NeighbVtx X → X ∈ V
4 3 pm4.71ri ⊢ N ∈ G NeighbVtx X ↔ X ∈ V ∧ N ∈ G NeighbVtx X
5 1 2 nbgrval ⊢ X ∈ V → G NeighbVtx X = n ∈ V ∖ X | ∃ e ∈ E X n ⊆ e
6 5 eleq2d ⊢ X ∈ V → N ∈ G NeighbVtx X ↔ N ∈ n ∈ V ∖ X | ∃ e ∈ E X n ⊆ e
7 preq2 ⊢ n = N → X n = X N
8 7 sseq1d ⊢ n = N → X n ⊆ e ↔ X N ⊆ e
9 8 rexbidv ⊢ n = N → ∃ e ∈ E X n ⊆ e ↔ ∃ e ∈ E X N ⊆ e
10 9 elrab ⊢ N ∈ n ∈ V ∖ X | ∃ e ∈ E X n ⊆ e ↔ N ∈ V ∖ X ∧ ∃ e ∈ E X N ⊆ e
11 eldifsn ⊢ N ∈ V ∖ X ↔ N ∈ V ∧ N ≠ X
12 11 anbi1i ⊢ N ∈ V ∖ X ∧ ∃ e ∈ E X N ⊆ e ↔ N ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e
13 10 12 bitri ⊢ N ∈ n ∈ V ∖ X | ∃ e ∈ E X n ⊆ e ↔ N ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e
14 6 13 bitrdi ⊢ X ∈ V → N ∈ G NeighbVtx X ↔ N ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e
15 14 pm5.32i ⊢ X ∈ V ∧ N ∈ G NeighbVtx X ↔ X ∈ V ∧ N ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e
16 df-3an ⊢ N ∈ V ∧ X ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e ↔ N ∈ V ∧ X ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e
17 anass ⊢ X ∈ V ∧ N ∈ V ∧ N ≠ X ↔ X ∈ V ∧ N ∈ V ∧ N ≠ X
18 ancom ⊢ X ∈ V ∧ N ∈ V ↔ N ∈ V ∧ X ∈ V
19 18 anbi1i ⊢ X ∈ V ∧ N ∈ V ∧ N ≠ X ↔ N ∈ V ∧ X ∈ V ∧ N ≠ X
20 17 19 bitr3i ⊢ X ∈ V ∧ N ∈ V ∧ N ≠ X ↔ N ∈ V ∧ X ∈ V ∧ N ≠ X
21 20 anbi1i ⊢ X ∈ V ∧ N ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e ↔ N ∈ V ∧ X ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e
22 anass ⊢ X ∈ V ∧ N ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e ↔ X ∈ V ∧ N ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e
23 16 21 22 3bitr2ri ⊢ X ∈ V ∧ N ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e ↔ N ∈ V ∧ X ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e
24 15 23 bitri ⊢ X ∈ V ∧ N ∈ G NeighbVtx X ↔ N ∈ V ∧ X ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e
25 4 24 bitri ⊢ N ∈ G NeighbVtx X ↔ N ∈ V ∧ X ∈ V ∧ N ≠ X ∧ ∃ e ∈ E X N ⊆ e