Metamath Proof Explorer


Theorem vtxdgval

Description: The degree of a vertex. (Contributed by Mario Carneiro, 12-Mar-2015) (Revised by Alexander van der Vekens, 20-Dec-2017) (Revised by AV, 10-Dec-2020) (Revised by AV, 22-Mar-2021)

Ref Expression
Hypotheses vtxdgval.v ⊢ V = Vtx ⁡ G
vtxdgval.i ⊢ I = iEdg ⁡ G
vtxdgval.a ⊢ A = dom ⁡ I
Assertion vtxdgval ⊢ U ∈ V → VtxDeg ⁡ G ⁡ U = x ∈ A | U ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = U

Proof

Step Hyp Ref Expression
1 vtxdgval.v ⊢ V = Vtx ⁡ G
2 vtxdgval.i ⊢ I = iEdg ⁡ G
3 vtxdgval.a ⊢ A = dom ⁡ I
4 1 1vgrex ⊢ U ∈ V → G ∈ V
5 1 2 3 vtxdgfval ⊢ G ∈ V → VtxDeg ⁡ G = u ∈ V ⟼ x ∈ A | u ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = u
6 4 5 syl ⊢ U ∈ V → VtxDeg ⁡ G = u ∈ V ⟼ x ∈ A | u ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = u
7 6 fveq1d ⊢ U ∈ V → VtxDeg ⁡ G ⁡ U = u ∈ V ⟼ x ∈ A | u ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = u ⁡ U
8 eleq1 ⊢ u = U → u ∈ I ⁡ x ↔ U ∈ I ⁡ x
9 8 rabbidv ⊢ u = U → x ∈ A | u ∈ I ⁡ x = x ∈ A | U ∈ I ⁡ x
10 9 fveq2d ⊢ u = U → x ∈ A | u ∈ I ⁡ x = x ∈ A | U ∈ I ⁡ x
11 sneq ⊢ u = U → u = U
12 11 eqeq2d ⊢ u = U → I ⁡ x = u ↔ I ⁡ x = U
13 12 rabbidv ⊢ u = U → x ∈ A | I ⁡ x = u = x ∈ A | I ⁡ x = U
14 13 fveq2d ⊢ u = U → x ∈ A | I ⁡ x = u = x ∈ A | I ⁡ x = U
15 10 14 oveq12d ⊢ u = U → x ∈ A | u ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = u = x ∈ A | U ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = U
16 eqid ⊢ u ∈ V ⟼ x ∈ A | u ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = u = u ∈ V ⟼ x ∈ A | u ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = u
17 ovex ⊢ x ∈ A | U ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = U ∈ V
18 15 16 17 fvmpt ⊢ U ∈ V → u ∈ V ⟼ x ∈ A | u ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = u ⁡ U = x ∈ A | U ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = U
19 7 18 eqtrd ⊢ U ∈ V → VtxDeg ⁡ G ⁡ U = x ∈ A | U ∈ I ⁡ x + 𝑒 x ∈ A | I ⁡ x = U