Metamath Proof Explorer


Theorem 4cyclusnfrgr

Description: A graph with a 4-cycle is not a friendhip graph. (Contributed by Alexander van der Vekens, 19-Dec-2017) (Revised by AV, 2-Apr-2021)

Ref Expression
Hypotheses 4cyclusnfrgr.v V = Vtx G
4cyclusnfrgr.e E = Edg G
Assertion 4cyclusnfrgr G USGraph A V C V A C B V D V B D A B E B C E C D E D A E G FriendGraph

Proof

Step Hyp Ref Expression
1 4cyclusnfrgr.v V = Vtx G
2 4cyclusnfrgr.e E = Edg G
3 simprl G USGraph A V C V A C B V D V B D A B E B C E C D E D A E A B E B C E
4 simprr G USGraph A V C V A C B V D V B D A B E B C E C D E D A E C D E D A E
5 simpl3 G USGraph A V C V A C B V D V B D A B E B C E C D E D A E B V D V B D
6 4cycl2vnunb A B E B C E C D E D A E B V D V B D ¬ ∃! x V A x x C E
7 3 4 5 6 syl3anc G USGraph A V C V A C B V D V B D A B E B C E C D E D A E ¬ ∃! x V A x x C E
8 1 2 frcond1 G FriendGraph A V C V A C ∃! x V A x x C E
9 pm2.24 ∃! x V A x x C E ¬ ∃! x V A x x C E ¬ G FriendGraph
10 8 9 syl6com A V C V A C G FriendGraph ¬ ∃! x V A x x C E ¬ G FriendGraph
11 10 3ad2ant2 G USGraph A V C V A C B V D V B D G FriendGraph ¬ ∃! x V A x x C E ¬ G FriendGraph
12 11 com23 G USGraph A V C V A C B V D V B D ¬ ∃! x V A x x C E G FriendGraph ¬ G FriendGraph
13 12 adantr G USGraph A V C V A C B V D V B D A B E B C E C D E D A E ¬ ∃! x V A x x C E G FriendGraph ¬ G FriendGraph
14 7 13 mpd G USGraph A V C V A C B V D V B D A B E B C E C D E D A E G FriendGraph ¬ G FriendGraph
15 14 pm2.01d G USGraph A V C V A C B V D V B D A B E B C E C D E D A E ¬ G FriendGraph
16 df-nel G FriendGraph ¬ G FriendGraph
17 15 16 sylibr G USGraph A V C V A C B V D V B D A B E B C E C D E D A E G FriendGraph
18 17 ex G USGraph A V C V A C B V D V B D A B E B C E C D E D A E G FriendGraph