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