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