Metamath Proof Explorer


Theorem isfrgr

Description: The property of being a friendship graph. (Contributed by Alexander van der Vekens, 4-Oct-2017) (Revised by AV, 29-Mar-2021) (Revised by AV, 3-Jan-2024)

Ref Expression
Hypotheses isfrgr.v V = Vtx G
isfrgr.e E = Edg G
Assertion isfrgr G FriendGraph G USGraph k V l V k ∃! x V x k x l E

Proof

Step Hyp Ref Expression
1 isfrgr.v V = Vtx G
2 isfrgr.e E = Edg G
3 fvex Vtx g V
4 fvex Edg g V
5 fveq2 g = G Vtx g = Vtx G
6 5 eqeq2d g = G v = Vtx g v = Vtx G
7 1 eqcomi Vtx G = V
8 7 eqeq2i v = Vtx G v = V
9 6 8 syl6bb g = G v = Vtx g v = V
10 fveq2 g = G Edg g = Edg G
11 10 eqeq2d g = G e = Edg g e = Edg G
12 2 eqcomi Edg G = E
13 12 eqeq2i e = Edg G e = E
14 11 13 syl6bb g = G e = Edg g e = E
15 9 14 anbi12d g = G v = Vtx g e = Edg g v = V e = E
16 simpl v = V e = E v = V
17 difeq1 v = V v k = V k
18 17 adantr v = V e = E v k = V k
19 reueq1 v = V ∃! x v x k x l e ∃! x V x k x l e
20 19 adantr v = V e = E ∃! x v x k x l e ∃! x V x k x l e
21 sseq2 e = E x k x l e x k x l E
22 21 adantl v = V e = E x k x l e x k x l E
23 22 reubidv v = V e = E ∃! x V x k x l e ∃! x V x k x l E
24 20 23 bitrd v = V e = E ∃! x v x k x l e ∃! x V x k x l E
25 18 24 raleqbidv v = V e = E l v k ∃! x v x k x l e l V k ∃! x V x k x l E
26 16 25 raleqbidv v = V e = E k v l v k ∃! x v x k x l e k V l V k ∃! x V x k x l E
27 15 26 syl6bi g = G v = Vtx g e = Edg g k v l v k ∃! x v x k x l e k V l V k ∃! x V x k x l E
28 3 4 27 sbc2iedv g = G [˙ Vtx g / v]˙ [˙ Edg g / e]˙ k v l v k ∃! x v x k x l e k V l V k ∃! x V x k x l E
29 df-frgr FriendGraph = g USGraph | [˙ Vtx g / v]˙ [˙ Edg g / e]˙ k v l v k ∃! x v x k x l e
30 28 29 elrab2 G FriendGraph G USGraph k V l V k ∃! x V x k x l E