Metamath Proof Explorer


Theorem frgrncvvdeqlem10

Description: Lemma 10 for frgrncvvdeq . (Contributed by Alexander van der Vekens, 24-Dec-2017) (Revised by AV, 10-May-2021) (Proof shortened by AV, 30-Dec-2021)

Ref Expression
Hypotheses frgrncvvdeq.v1 V = Vtx G
frgrncvvdeq.e E = Edg G
frgrncvvdeq.nx D = G NeighbVtx X
frgrncvvdeq.ny N = G NeighbVtx Y
frgrncvvdeq.x φ X V
frgrncvvdeq.y φ Y V
frgrncvvdeq.ne φ X Y
frgrncvvdeq.xy φ Y D
frgrncvvdeq.f φ G FriendGraph
frgrncvvdeq.a A = x D ι y N | x y E
Assertion frgrncvvdeqlem10 φ A : D 1-1 onto N

Proof

Step Hyp Ref Expression
1 frgrncvvdeq.v1 V = Vtx G
2 frgrncvvdeq.e E = Edg G
3 frgrncvvdeq.nx D = G NeighbVtx X
4 frgrncvvdeq.ny N = G NeighbVtx Y
5 frgrncvvdeq.x φ X V
6 frgrncvvdeq.y φ Y V
7 frgrncvvdeq.ne φ X Y
8 frgrncvvdeq.xy φ Y D
9 frgrncvvdeq.f φ G FriendGraph
10 frgrncvvdeq.a A = x D ι y N | x y E
11 1 2 3 4 5 6 7 8 9 10 frgrncvvdeqlem8 φ A : D 1-1 N
12 1 2 3 4 5 6 7 8 9 10 frgrncvvdeqlem9 φ A : D onto N
13 df-f1o A : D 1-1 onto N A : D 1-1 N A : D onto N
14 11 12 13 sylanbrc φ A : D 1-1 onto N