Metamath Proof Explorer


Theorem frcond1

Description: The friendship condition: any two (different) vertices in a friendship graph have a unique common neighbor. (Contributed by Alexander van der Vekens, 19-Dec-2017) (Revised by AV, 29-Mar-2021)

Ref Expression
Hypotheses frcond1.v V = Vtx G
frcond1.e E = Edg G
Assertion frcond1 G FriendGraph A V C V A C ∃! b V A b b C E

Proof

Step Hyp Ref Expression
1 frcond1.v V = Vtx G
2 frcond1.e E = Edg G
3 1 2 isfrgr G FriendGraph G USGraph k V l V k ∃! b V b k b l E
4 preq2 k = A b k = b A
5 4 preq1d k = A b k b l = b A b l
6 5 sseq1d k = A b k b l E b A b l E
7 6 reubidv k = A ∃! b V b k b l E ∃! b V b A b l E
8 preq2 l = C b l = b C
9 8 preq2d l = C b A b l = b A b C
10 9 sseq1d l = C b A b l E b A b C E
11 10 reubidv l = C ∃! b V b A b l E ∃! b V b A b C E
12 simp1 A V C V A C A V
13 sneq k = A k = A
14 13 difeq2d k = A V k = V A
15 14 adantl A V C V A C k = A V k = V A
16 necom A C C A
17 16 biimpi A C C A
18 17 anim2i C V A C C V C A
19 18 3adant1 A V C V A C C V C A
20 eldifsn C V A C V C A
21 19 20 sylibr A V C V A C C V A
22 7 11 12 15 21 rspc2vd A V C V A C k V l V k ∃! b V b k b l E ∃! b V b A b C E
23 prcom b A = A b
24 23 preq1i b A b C = A b b C
25 24 sseq1i b A b C E A b b C E
26 25 reubii ∃! b V b A b C E ∃! b V A b b C E
27 26 biimpi ∃! b V b A b C E ∃! b V A b b C E
28 22 27 syl6com k V l V k ∃! b V b k b l E A V C V A C ∃! b V A b b C E
29 3 28 simplbiim G FriendGraph A V C V A C ∃! b V A b b C E