Metamath Proof Explorer


Theorem 0prjspnrel

Description: In the zero-dimensional projective space, all vectors are equivalent to the unit vector. (Contributed by Steven Nguyen, 7-Jun-2023)

Ref Expression
Hypotheses 0prjspnrel.e ˙ = x y | x B y B l S x = l · ˙ y
0prjspnrel.b B = Base W 0 W
0prjspnrel.x · ˙ = W
0prjspnrel.s S = Base K
0prjspnrel.w W = K freeLMod 0 0
0prjspnrel.1 1 ˙ = K unitVec 0 0 0
Assertion 0prjspnrel K DivRing X B X ˙ 1 ˙

Proof

Step Hyp Ref Expression
1 0prjspnrel.e ˙ = x y | x B y B l S x = l · ˙ y
2 0prjspnrel.b B = Base W 0 W
3 0prjspnrel.x · ˙ = W
4 0prjspnrel.s S = Base K
5 0prjspnrel.w W = K freeLMod 0 0
6 0prjspnrel.1 1 ˙ = K unitVec 0 0 0
7 simpr K DivRing X B X B
8 2 5 6 0prjspnlem K DivRing 1 ˙ B
9 8 adantr K DivRing X B 1 ˙ B
10 ovexd K DivRing X B 0 0 V
11 difss Base W 0 W Base W
12 2 11 eqsstri B Base W
13 12 sseli X B X Base W
14 13 adantl K DivRing X B X Base W
15 eqid Base K = Base K
16 eqid Base W = Base W
17 5 15 16 frlmbasf 0 0 V X Base W X : 0 0 Base K
18 10 14 17 syl2anc K DivRing X B X : 0 0 Base K
19 c0ex 0 V
20 19 snid 0 0
21 fz0sn 0 0 = 0
22 20 21 eleqtrri 0 0 0
23 22 a1i K DivRing X B 0 0 0
24 18 23 ffvelrnd K DivRing X B X 0 Base K
25 sneq n = X 0 n = X 0
26 25 xpeq2d n = X 0 0 0 × n = 0 0 × X 0
27 26 eqeq2d n = X 0 X = 0 0 × n X = 0 0 × X 0
28 27 adantl K DivRing X B n = X 0 X = 0 0 × n X = 0 0 × X 0
29 5 15 16 frlmbasmap 0 0 V X Base W X Base K 0 0
30 10 14 29 syl2anc K DivRing X B X Base K 0 0
31 fvex Base K V
32 21 31 19 mapsnconst X Base K 0 0 X = 0 0 × X 0
33 30 32 syl K DivRing X B X = 0 0 × X 0
34 24 28 33 rspcedvd K DivRing X B n Base K X = 0 0 × n
35 simprl K DivRing X B n Base K X = 0 0 × n n Base K
36 35 4 eleqtrrdi K DivRing X B n Base K X = 0 0 × n n S
37 oveq1 m = n m · ˙ 1 ˙ = n · ˙ 1 ˙
38 37 eqeq2d m = n X = m · ˙ 1 ˙ X = n · ˙ 1 ˙
39 38 adantl K DivRing X B n Base K X = 0 0 × n m = n X = m · ˙ 1 ˙ X = n · ˙ 1 ˙
40 ovexd K DivRing X B n Base K 0 0 V
41 simpr K DivRing X B n Base K n Base K
42 8 ad2antrr K DivRing X B n Base K 1 ˙ B
43 12 42 sselid K DivRing X B n Base K 1 ˙ Base W
44 eqid K = K
45 5 16 15 40 41 43 3 44 frlmvscafval K DivRing X B n Base K n · ˙ 1 ˙ = 0 0 × n K f 1 ˙
46 5 15 16 frlmbasf 0 0 V 1 ˙ Base W 1 ˙ : 0 0 Base K
47 40 43 46 syl2anc K DivRing X B n Base K 1 ˙ : 0 0 Base K
48 drngring K DivRing K Ring
49 eqid 1 K = 1 K
50 15 49 ringidcl K Ring 1 K Base K
51 48 50 syl K DivRing 1 K Base K
52 51 ad2antrr K DivRing X B n Base K 1 K Base K
53 52 snssd K DivRing X B n Base K 1 K Base K
54 6 a1i d 0 0 1 ˙ = K unitVec 0 0 0
55 elfz1eq d 0 0 d = 0
56 54 55 fveq12d d 0 0 1 ˙ d = K unitVec 0 0 0 0
57 56 adantl K DivRing X B n Base K d 0 0 1 ˙ d = K unitVec 0 0 0 0
58 eqid K unitVec 0 0 = K unitVec 0 0
59 simplll K DivRing X B n Base K d 0 0 K DivRing
60 ovexd K DivRing X B n Base K d 0 0 0 0 V
61 22 a1i K DivRing X B n Base K d 0 0 0 0 0
62 58 59 60 61 49 uvcvv1 K DivRing X B n Base K d 0 0 K unitVec 0 0 0 0 = 1 K
63 fvex K unitVec 0 0 0 0 V
64 63 elsn K unitVec 0 0 0 0 1 K K unitVec 0 0 0 0 = 1 K
65 62 64 sylibr K DivRing X B n Base K d 0 0 K unitVec 0 0 0 0 1 K
66 57 65 eqeltrd K DivRing X B n Base K d 0 0 1 ˙ d 1 K
67 66 ralrimiva K DivRing X B n Base K d 0 0 1 ˙ d 1 K
68 frnssb 1 K Base K d 0 0 1 ˙ d 1 K 1 ˙ : 0 0 Base K 1 ˙ : 0 0 1 K
69 53 67 68 syl2anc K DivRing X B n Base K 1 ˙ : 0 0 Base K 1 ˙ : 0 0 1 K
70 47 69 mpbid K DivRing X B n Base K 1 ˙ : 0 0 1 K
71 vex n V
72 71 a1i K DivRing X B n Base K n V
73 elsni c 1 K c = 1 K
74 73 oveq2d c 1 K n K c = n K 1 K
75 48 ad2antrr K DivRing X B n Base K K Ring
76 15 44 49 ringridm K Ring n Base K n K 1 K = n
77 75 41 76 syl2anc K DivRing X B n Base K n K 1 K = n
78 74 77 sylan9eqr K DivRing X B n Base K c 1 K n K c = n
79 40 70 72 72 78 caofid2 K DivRing X B n Base K 0 0 × n K f 1 ˙ = 0 0 × n
80 45 79 eqtrd K DivRing X B n Base K n · ˙ 1 ˙ = 0 0 × n
81 80 eqeq2d K DivRing X B n Base K X = n · ˙ 1 ˙ X = 0 0 × n
82 81 biimprd K DivRing X B n Base K X = 0 0 × n X = n · ˙ 1 ˙
83 82 impr K DivRing X B n Base K X = 0 0 × n X = n · ˙ 1 ˙
84 36 39 83 rspcedvd K DivRing X B n Base K X = 0 0 × n m S X = m · ˙ 1 ˙
85 34 84 rexlimddv K DivRing X B m S X = m · ˙ 1 ˙
86 1 prjsprel X ˙ 1 ˙ X B 1 ˙ B m S X = m · ˙ 1 ˙
87 7 9 85 86 syl21anbrc K DivRing X B X ˙ 1 ˙