Metamath Proof Explorer


Theorem frgrwopreg

Description: In a friendship graph there are either no vertices ( A = (/) ) or exactly one vertex ( ( #A ) = 1 ) having degree K , or all ( B = (/) ) or all except one vertices ( ( #B ) = 1 ) have degree K . (Contributed by Alexander van der Vekens, 31-Dec-2017) (Revised by AV, 10-May-2021) (Proof shortened by AV, 3-Jan-2022)

Ref Expression
Hypotheses frgrwopreg.v V=VtxG
frgrwopreg.d D=VtxDegG
frgrwopreg.a A=xV|Dx=K
frgrwopreg.b B=VA
Assertion frgrwopreg GFriendGraphA=1A=B=1B=

Proof

Step Hyp Ref Expression
1 frgrwopreg.v V=VtxG
2 frgrwopreg.d D=VtxDegG
3 frgrwopreg.a A=xV|Dx=K
4 frgrwopreg.b B=VA
5 1 2 3 4 frgrwopreglem1 AVBV
6 hashv01gt1 AVA=0A=11<A
7 hasheq0 AVA=0A=
8 biidd AVA=1A=1
9 biidd AV1<A1<A
10 7 8 9 3orbi123d AVA=0A=11<AA=A=11<A
11 hashv01gt1 BVB=0B=11<B
12 hasheq0 BVB=0B=
13 biidd BVB=1B=1
14 biidd BV1<B1<B
15 12 13 14 3orbi123d BVB=0B=11<BB=B=11<B
16 olc B=B=1B=
17 16 olcd B=A=1A=B=1B=
18 17 2a1d B=A=A=11<AGFriendGraphA=1A=B=1B=
19 orc B=1B=1B=
20 19 olcd B=1A=1A=B=1B=
21 20 2a1d B=1A=A=11<AGFriendGraphA=1A=B=1B=
22 olc A=A=1A=
23 22 orcd A=A=1A=B=1B=
24 23 2a1d A=1<BGFriendGraphA=1A=B=1B=
25 orc A=1A=1A=
26 25 orcd A=1A=1A=B=1B=
27 26 2a1d A=11<BGFriendGraphA=1A=B=1B=
28 eqid EdgG=EdgG
29 1 2 3 4 28 frgrwopreglem5 GFriendGraph1<A1<BaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgG
30 frgrusgr GFriendGraphGUSGraph
31 simplll GUSGraphaAxAbByBaxbyGUSGraph
32 elrabi axV|Dx=KaV
33 32 3 eleq2s aAaV
34 33 adantr aAxAaV
35 34 ad3antlr GUSGraphaAxAbByBaxbyaV
36 rabidim1 xxV|Dx=KxV
37 36 3 eleq2s xAxV
38 37 adantl aAxAxV
39 38 ad3antlr GUSGraphaAxAbByBaxbyxV
40 simprl GUSGraphaAxAbByBaxbyax
41 eldifi bVAbV
42 41 4 eleq2s bBbV
43 42 adantr bByBbV
44 43 ad2antlr GUSGraphaAxAbByBaxbybV
45 eldifi yVAyV
46 45 4 eleq2s yByV
47 46 adantl bByByV
48 47 ad2antlr GUSGraphaAxAbByBaxbyyV
49 simprr GUSGraphaAxAbByBaxbyby
50 1 28 4cyclusnfrgr GUSGraphaVxVaxbVyVbyabEdgGbxEdgGxyEdgGyaEdgGGFriendGraph
51 31 35 39 40 44 48 49 50 syl133anc GUSGraphaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgGGFriendGraph
52 51 exp4b GUSGraphaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgGGFriendGraph
53 52 3impd GUSGraphaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgGGFriendGraph
54 df-nel GFriendGraph¬GFriendGraph
55 pm2.21 ¬GFriendGraphGFriendGraphA=1A=B=1B=
56 54 55 sylbi GFriendGraphGFriendGraphA=1A=B=1B=
57 53 56 syl6 GUSGraphaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgGGFriendGraphA=1A=B=1B=
58 57 rexlimdvva GUSGraphaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgGGFriendGraphA=1A=B=1B=
59 58 rexlimdvva GUSGraphaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgGGFriendGraphA=1A=B=1B=
60 59 com23 GUSGraphGFriendGraphaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgGA=1A=B=1B=
61 30 60 mpcom GFriendGraphaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgGA=1A=B=1B=
62 61 3ad2ant1 GFriendGraph1<A1<BaAxAbByBaxbyabEdgGbxEdgGxyEdgGyaEdgGA=1A=B=1B=
63 29 62 mpd GFriendGraph1<A1<BA=1A=B=1B=
64 63 3exp GFriendGraph1<A1<BA=1A=B=1B=
65 64 com3l 1<A1<BGFriendGraphA=1A=B=1B=
66 24 27 65 3jaoi A=A=11<A1<BGFriendGraphA=1A=B=1B=
67 66 com12 1<BA=A=11<AGFriendGraphA=1A=B=1B=
68 18 21 67 3jaoi B=B=11<BA=A=11<AGFriendGraphA=1A=B=1B=
69 15 68 syl6bi BVB=0B=11<BA=A=11<AGFriendGraphA=1A=B=1B=
70 11 69 mpd BVA=A=11<AGFriendGraphA=1A=B=1B=
71 70 com12 A=A=11<ABVGFriendGraphA=1A=B=1B=
72 10 71 syl6bi AVA=0A=11<ABVGFriendGraphA=1A=B=1B=
73 6 72 mpd AVBVGFriendGraphA=1A=B=1B=
74 73 imp AVBVGFriendGraphA=1A=B=1B=
75 5 74 ax-mp GFriendGraphA=1A=B=1B=