Metamath Proof Explorer


Theorem nbuhgr2vtx1edgblem

Description: Lemma for nbuhgr2vtx1edgb . This reverse direction of nbgr2vtx1edg only holds for classes whose edges are subsets of the set of vertices, which is the property of hypergraphs. (Contributed by AV, 2-Nov-2020) (Proof shortened by AV, 13-Feb-2022)

Ref Expression
Hypotheses nbgr2vtx1edg.v V = Vtx G
nbgr2vtx1edg.e E = Edg G
Assertion nbuhgr2vtx1edgblem G UHGraph V = a b a G NeighbVtx b a b E

Proof

Step Hyp Ref Expression
1 nbgr2vtx1edg.v V = Vtx G
2 nbgr2vtx1edg.e E = Edg G
3 1 2 nbgrel a G NeighbVtx b a V b V a b e E b a e
4 2 eleq2i e E e Edg G
5 edguhgr G UHGraph e Edg G e 𝒫 Vtx G
6 4 5 sylan2b G UHGraph e E e 𝒫 Vtx G
7 1 eqeq1i V = a b Vtx G = a b
8 pweq Vtx G = a b 𝒫 Vtx G = 𝒫 a b
9 8 eleq2d Vtx G = a b e 𝒫 Vtx G e 𝒫 a b
10 velpw e 𝒫 a b e a b
11 9 10 bitrdi Vtx G = a b e 𝒫 Vtx G e a b
12 7 11 sylbi V = a b e 𝒫 Vtx G e a b
13 12 adantl G UHGraph e E V = a b e 𝒫 Vtx G e a b
14 prcom b a = a b
15 14 sseq1i b a e a b e
16 eqss a b = e a b e e a b
17 eleq1a e E a b = e a b E
18 17 a1i a V b V a b e E a b = e a b E
19 18 com13 a b = e e E a V b V a b a b E
20 16 19 sylbir a b e e a b e E a V b V a b a b E
21 20 ex a b e e a b e E a V b V a b a b E
22 15 21 sylbi b a e e a b e E a V b V a b a b E
23 22 com13 e E e a b b a e a V b V a b a b E
24 23 ad2antlr G UHGraph e E V = a b e a b b a e a V b V a b a b E
25 13 24 sylbid G UHGraph e E V = a b e 𝒫 Vtx G b a e a V b V a b a b E
26 25 ex G UHGraph e E V = a b e 𝒫 Vtx G b a e a V b V a b a b E
27 6 26 mpid G UHGraph e E V = a b b a e a V b V a b a b E
28 27 impancom G UHGraph V = a b e E b a e a V b V a b a b E
29 28 com14 a V b V a b e E b a e G UHGraph V = a b a b E
30 29 rexlimdv a V b V a b e E b a e G UHGraph V = a b a b E
31 30 3impia a V b V a b e E b a e G UHGraph V = a b a b E
32 31 com12 G UHGraph V = a b a V b V a b e E b a e a b E
33 3 32 syl5bi G UHGraph V = a b a G NeighbVtx b a b E
34 33 3impia G UHGraph V = a b a G NeighbVtx b a b E