Metamath Proof Explorer


Theorem 3cyclfrgr

Description: Every vertex in a friendship graph (with more than 1 vertex) is part of a 3-cycle. (Contributed by Alexander van der Vekens, 19-Nov-2017) (Revised by AV, 2-Apr-2021)

Ref Expression
Hypothesis 3cyclfrgr.v V = Vtx G
Assertion 3cyclfrgr G FriendGraph 1 < V v V f p f Cycles G p f = 3 p 0 = v

Proof

Step Hyp Ref Expression
1 3cyclfrgr.v V = Vtx G
2 eqid Edg G = Edg G
3 1 2 3cyclfrgrrn G FriendGraph 1 < V v V b V c V v b Edg G b c Edg G c v Edg G
4 frgrusgr G FriendGraph G USGraph
5 usgrumgr G USGraph G UMGraph
6 4 5 syl G FriendGraph G UMGraph
7 6 ad4antr G FriendGraph 1 < V v V b V c V v b Edg G b c Edg G c v Edg G G UMGraph
8 simpr G FriendGraph 1 < V v V v V
9 8 anim1i G FriendGraph 1 < V v V b V c V v V b V c V
10 3anass v V b V c V v V b V c V
11 9 10 sylibr G FriendGraph 1 < V v V b V c V v V b V c V
12 11 adantr G FriendGraph 1 < V v V b V c V v b Edg G b c Edg G c v Edg G v V b V c V
13 simpr G FriendGraph 1 < V v V b V c V v b Edg G b c Edg G c v Edg G v b Edg G b c Edg G c v Edg G
14 1 2 umgr3cyclex G UMGraph v V b V c V v b Edg G b c Edg G c v Edg G f p f Cycles G p f = 3 p 0 = v
15 7 12 13 14 syl3anc G FriendGraph 1 < V v V b V c V v b Edg G b c Edg G c v Edg G f p f Cycles G p f = 3 p 0 = v
16 15 ex G FriendGraph 1 < V v V b V c V v b Edg G b c Edg G c v Edg G f p f Cycles G p f = 3 p 0 = v
17 16 rexlimdvva G FriendGraph 1 < V v V b V c V v b Edg G b c Edg G c v Edg G f p f Cycles G p f = 3 p 0 = v
18 17 ralimdva G FriendGraph 1 < V v V b V c V v b Edg G b c Edg G c v Edg G v V f p f Cycles G p f = 3 p 0 = v
19 3 18 mpd G FriendGraph 1 < V v V f p f Cycles G p f = 3 p 0 = v