Metamath Proof Explorer


Theorem cyclnumvtx

Description: The number of vertices of a (non-trivial) cycle is the number of edges in the cycle. (Contributed by AV, 5-Oct-2025)

Ref Expression
Assertion cyclnumvtx 1 F F Cycles G P ran P = F

Proof

Step Hyp Ref Expression
1 iscycl F Cycles G P F Paths G P P 0 = P F
2 pthiswlk F Paths G P F Walks G P
3 eqid Vtx G = Vtx G
4 3 wlkp F Walks G P P : 0 F Vtx G
5 wlkcl F Walks G P F 0
6 elnnnn0c F F 0 1 F
7 fdm P : 0 F Vtx G dom P = 0 F
8 7 3ad2ant1 P : 0 F Vtx G F P 0 = P F dom P = 0 F
9 8 difeq1d P : 0 F Vtx G F P 0 = P F dom P 1 F = 0 F 1 F
10 nnnn0 F F 0
11 fz0sn0fz1 F 0 0 F = 0 1 F
12 10 11 syl F 0 F = 0 1 F
13 12 difeq1d F 0 F 1 F = 0 1 F 1 F
14 1e0p1 1 = 0 + 1
15 14 oveq1i 1 F = 0 + 1 F
16 15 ineq2i 0 1 F = 0 0 + 1 F
17 elnn0uz F 0 F 0
18 10 17 sylib F F 0
19 fzpreddisj F 0 0 0 + 1 F =
20 18 19 syl F 0 0 + 1 F =
21 16 20 eqtrid F 0 1 F =
22 undif5 0 1 F = 0 1 F 1 F = 0
23 21 22 syl F 0 1 F 1 F = 0
24 13 23 eqtrd F 0 F 1 F = 0
25 24 3ad2ant2 P : 0 F Vtx G F P 0 = P F 0 F 1 F = 0
26 9 25 eqtrd P : 0 F Vtx G F P 0 = P F dom P 1 F = 0
27 26 imaeq2d P : 0 F Vtx G F P 0 = P F P dom P 1 F = P 0
28 ffn P : 0 F Vtx G P Fn 0 F
29 0elfz F 0 0 0 F
30 10 29 syl F 0 0 F
31 28 30 anim12i P : 0 F Vtx G F P Fn 0 F 0 0 F
32 31 3adant3 P : 0 F Vtx G F P 0 = P F P Fn 0 F 0 0 F
33 fnsnfv P Fn 0 F 0 0 F P 0 = P 0
34 32 33 syl P : 0 F Vtx G F P 0 = P F P 0 = P 0
35 27 34 eqtr4d P : 0 F Vtx G F P 0 = P F P dom P 1 F = P 0
36 elfz1end F F 1 F
37 36 biimpi F F 1 F
38 37 3ad2ant2 P : 0 F Vtx G F P 0 = P F F 1 F
39 38 fvresd P : 0 F Vtx G F P 0 = P F P 1 F F = P F
40 ffun P : 0 F Vtx G Fun P
41 40 funresd P : 0 F Vtx G Fun P 1 F
42 41 3ad2ant1 P : 0 F Vtx G F P 0 = P F Fun P 1 F
43 fz1ssfz0 1 F 0 F
44 43 7 sseqtrrid P : 0 F Vtx G 1 F dom P
45 44 3ad2ant1 P : 0 F Vtx G F P 0 = P F 1 F dom P
46 ssdmres 1 F dom P dom P 1 F = 1 F
47 45 46 sylib P : 0 F Vtx G F P 0 = P F dom P 1 F = 1 F
48 38 47 eleqtrrd P : 0 F Vtx G F P 0 = P F F dom P 1 F
49 fvelrn Fun P 1 F F dom P 1 F P 1 F F ran P 1 F
50 42 48 49 syl2anc P : 0 F Vtx G F P 0 = P F P 1 F F ran P 1 F
51 39 50 eqeltrrd P : 0 F Vtx G F P 0 = P F P F ran P 1 F
52 eleq1 P 0 = P F P 0 ran P 1 F P F ran P 1 F
53 52 3ad2ant3 P : 0 F Vtx G F P 0 = P F P 0 ran P 1 F P F ran P 1 F
54 51 53 mpbird P : 0 F Vtx G F P 0 = P F P 0 ran P 1 F
55 54 snssd P : 0 F Vtx G F P 0 = P F P 0 ran P 1 F
56 35 55 eqsstrd P : 0 F Vtx G F P 0 = P F P dom P 1 F ran P 1 F
57 56 3exp P : 0 F Vtx G F P 0 = P F P dom P 1 F ran P 1 F
58 57 com3l F P 0 = P F P : 0 F Vtx G P dom P 1 F ran P 1 F
59 6 58 sylbir F 0 1 F P 0 = P F P : 0 F Vtx G P dom P 1 F ran P 1 F
60 59 expcom 1 F F 0 P 0 = P F P : 0 F Vtx G P dom P 1 F ran P 1 F
61 60 com14 P : 0 F Vtx G F 0 P 0 = P F 1 F P dom P 1 F ran P 1 F
62 4 5 61 sylc F Walks G P P 0 = P F 1 F P dom P 1 F ran P 1 F
63 2 62 syl F Paths G P P 0 = P F 1 F P dom P 1 F ran P 1 F
64 63 imp F Paths G P P 0 = P F 1 F P dom P 1 F ran P 1 F
65 1 64 sylbi F Cycles G P 1 F P dom P 1 F ran P 1 F
66 65 impcom 1 F F Cycles G P P dom P 1 F ran P 1 F
67 imadifssran P dom P 1 F ran P 1 F ran P = ran P 1 F
68 67 fveq2d P dom P 1 F ran P 1 F ran P = ran P 1 F
69 66 68 syl 1 F F Cycles G P ran P = ran P 1 F
70 cyclispth F Cycles G P F Paths G P
71 pthdifv F Paths G P P 1 F : 1 F 1-1 Vtx G
72 40 adantl F 0 P : 0 F Vtx G Fun P
73 fzfid F 0 0 F Fin
74 fnfi P Fn 0 F 0 F Fin P Fin
75 28 73 74 syl2anr F 0 P : 0 F Vtx G P Fin
76 1eluzge0 1 0
77 fzss1 1 0 1 F 0 F
78 76 77 mp1i F 0 P : 0 F Vtx G 1 F 0 F
79 7 adantl F 0 P : 0 F Vtx G dom P = 0 F
80 78 79 sseqtrrd F 0 P : 0 F Vtx G 1 F dom P
81 72 75 80 3jca F 0 P : 0 F Vtx G Fun P P Fin 1 F dom P
82 5 4 81 syl2anc F Walks G P Fun P P Fin 1 F dom P
83 2 82 syl F Paths G P Fun P P Fin 1 F dom P
84 83 adantr F Paths G P P 1 F : 1 F 1-1 Vtx G Fun P P Fin 1 F dom P
85 hashres Fun P P Fin 1 F dom P P 1 F = 1 F
86 84 85 syl F Paths G P P 1 F : 1 F 1-1 Vtx G P 1 F = 1 F
87 ovexd F Paths G P P 1 F : 1 F 1-1 Vtx G 1 F V
88 hashf1rn 1 F V P 1 F : 1 F 1-1 Vtx G P 1 F = ran P 1 F
89 87 88 sylancom F Paths G P P 1 F : 1 F 1-1 Vtx G P 1 F = ran P 1 F
90 2 5 syl F Paths G P F 0
91 hashfz1 F 0 1 F = F
92 90 91 syl F Paths G P 1 F = F
93 92 adantr F Paths G P P 1 F : 1 F 1-1 Vtx G 1 F = F
94 86 89 93 3eqtr3d F Paths G P P 1 F : 1 F 1-1 Vtx G ran P 1 F = F
95 70 71 94 syl2anc2 F Cycles G P ran P 1 F = F
96 95 adantl 1 F F Cycles G P ran P 1 F = F
97 69 96 eqtrd 1 F F Cycles G P ran P = F