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Metamath Proof Explorer

Theorem 1hegrvtxdg1r

Description: The vertex degree of a graph with one hyperedge, case 3: an edge from some other vertex to the given vertex contributes one to the vertex's degree. (Contributed by Mario Carneiro, 12-Mar-2015) (Revised by Alexander van der Vekens, 22-Dec-2017) (Revised by AV, 23-Feb-2021)

Ref Expression
Hypotheses 1hegrvtxdg1.a ⊢ φ → A ∈ X
1hegrvtxdg1.b ⊢ φ → B ∈ V
1hegrvtxdg1.c ⊢ φ → C ∈ V
1hegrvtxdg1.n ⊢ φ → B ≠ C
1hegrvtxdg1.x ⊢ φ → E ∈ 𝒫 V
1hegrvtxdg1.i ⊢ φ → iEdg ⁡ G = A E
1hegrvtxdg1.e ⊢ φ → B C ⊆ E
1hegrvtxdg1.v ⊢ φ → Vtx ⁡ G = V
Assertion 1hegrvtxdg1r ⊢ φ → VtxDeg ⁡ G ⁡ C = 1

Proof

Step Hyp Ref Expression
1 1hegrvtxdg1.a ⊢ φ → A ∈ X
2 1hegrvtxdg1.b ⊢ φ → B ∈ V
3 1hegrvtxdg1.c ⊢ φ → C ∈ V
4 1hegrvtxdg1.n ⊢ φ → B ≠ C
5 1hegrvtxdg1.x ⊢ φ → E ∈ 𝒫 V
6 1hegrvtxdg1.i ⊢ φ → iEdg ⁡ G = A E
7 1hegrvtxdg1.e ⊢ φ → B C ⊆ E
8 1hegrvtxdg1.v ⊢ φ → Vtx ⁡ G = V
9 4 necomd ⊢ φ → C ≠ B
10 prcom ⊢ C B = B C
11 10 7 eqsstrid ⊢ φ → C B ⊆ E
12 1 3 2 9 5 6 11 8 1hegrvtxdg1 ⊢ φ → VtxDeg ⁡ G ⁡ C = 1