Metamath Proof Explorer


Theorem 1egrvtxdg1

Description: The vertex degree of a one-edge graph, case 2: an edge from the given vertex to some other 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, 21-Feb-2021)

Ref Expression
Hypotheses 1egrvtxdg1.v φ Vtx G = V
1egrvtxdg1.a φ A X
1egrvtxdg1.b φ B V
1egrvtxdg1.c φ C V
1egrvtxdg1.n φ B C
1egrvtxdg1.i φ iEdg G = A B C
Assertion 1egrvtxdg1 φ VtxDeg G B = 1

Proof

Step Hyp Ref Expression
1 1egrvtxdg1.v φ Vtx G = V
2 1egrvtxdg1.a φ A X
3 1egrvtxdg1.b φ B V
4 1egrvtxdg1.c φ C V
5 1egrvtxdg1.n φ B C
6 1egrvtxdg1.i φ iEdg G = A B C
7 eqid Vtx G = Vtx G
8 3 1 eleqtrrd φ B Vtx G
9 4 1 eleqtrrd φ C Vtx G
10 7 2 8 9 6 5 usgr1e φ G USGraph
11 eqid iEdg G = iEdg G
12 eqid dom iEdg G = dom iEdg G
13 eqid VtxDeg G = VtxDeg G
14 7 11 12 13 vtxdusgrval G USGraph B Vtx G VtxDeg G B = x dom iEdg G | B iEdg G x
15 10 8 14 syl2anc φ VtxDeg G B = x dom iEdg G | B iEdg G x
16 dmeq iEdg G = A B C dom iEdg G = dom A B C
17 16 adantl φ iEdg G = A B C dom iEdg G = dom A B C
18 prex B C V
19 dmsnopg B C V dom A B C = A
20 18 19 mp1i φ iEdg G = A B C dom A B C = A
21 17 20 eqtrd φ iEdg G = A B C dom iEdg G = A
22 fveq1 iEdg G = A B C iEdg G x = A B C x
23 22 eleq2d iEdg G = A B C B iEdg G x B A B C x
24 23 adantl φ iEdg G = A B C B iEdg G x B A B C x
25 21 24 rabeqbidv φ iEdg G = A B C x dom iEdg G | B iEdg G x = x A | B A B C x
26 25 fveq2d φ iEdg G = A B C x dom iEdg G | B iEdg G x = x A | B A B C x
27 fveq2 x = A A B C x = A B C A
28 27 eleq2d x = A B A B C x B A B C A
29 28 rabsnif x A | B A B C x = if B A B C A A
30 prid1g B V B B C
31 3 30 syl φ B B C
32 fvsng A X B C V A B C A = B C
33 2 18 32 sylancl φ A B C A = B C
34 31 33 eleqtrrd φ B A B C A
35 34 iftrued φ if B A B C A A = A
36 29 35 syl5eq φ x A | B A B C x = A
37 36 fveq2d φ x A | B A B C x = A
38 hashsng A X A = 1
39 2 38 syl φ A = 1
40 37 39 eqtrd φ x A | B A B C x = 1
41 40 adantr φ iEdg G = A B C x A | B A B C x = 1
42 26 41 eqtrd φ iEdg G = A B C x dom iEdg G | B iEdg G x = 1
43 6 42 mpdan φ x dom iEdg G | B iEdg G x = 1
44 15 43 eqtrd φ VtxDeg G B = 1