Metamath Proof Explorer


Theorem lvecdim0

Description: A vector space of dimension zero is reduced to its identity element, biconditional version. (Contributed by Thierry Arnoux, 31-Jul-2023)

Ref Expression
Hypothesis lvecdim0.1 0 ˙ = 0 V
Assertion lvecdim0 V LVec dim V = 0 Base V = 0 ˙

Proof

Step Hyp Ref Expression
1 lvecdim0.1 0 ˙ = 0 V
2 1 lvecdim0i V LVec dim V = 0 Base V = 0 ˙
3 simpl V LVec Base V = 0 ˙ V LVec
4 eqid LBasis V = LBasis V
5 4 lbsex V LVec LBasis V
6 n0 LBasis V b b LBasis V
7 5 6 sylib V LVec b b LBasis V
8 3 7 syl V LVec Base V = 0 ˙ b b LBasis V
9 1 fvexi 0 ˙ V
10 9 snid 0 ˙ 0 ˙
11 simpr V LVec Base V = 0 ˙ b LBasis V b = 0 ˙ b = 0 ˙
12 10 11 eleqtrrid V LVec Base V = 0 ˙ b LBasis V b = 0 ˙ 0 ˙ b
13 simplll V LVec Base V = 0 ˙ b LBasis V b = 0 ˙ V LVec
14 4 lbslinds LBasis V LIndS V
15 simplr V LVec Base V = 0 ˙ b LBasis V b = 0 ˙ b LBasis V
16 14 15 sseldi V LVec Base V = 0 ˙ b LBasis V b = 0 ˙ b LIndS V
17 1 0nellinds V LVec b LIndS V ¬ 0 ˙ b
18 13 16 17 syl2anc V LVec Base V = 0 ˙ b LBasis V b = 0 ˙ ¬ 0 ˙ b
19 12 18 pm2.65da V LVec Base V = 0 ˙ b LBasis V ¬ b = 0 ˙
20 simpr V LVec Base V = 0 ˙ b LBasis V b LBasis V
21 eqid Base V = Base V
22 21 4 lbsss b LBasis V b Base V
23 20 22 syl V LVec Base V = 0 ˙ b LBasis V b Base V
24 simplr V LVec Base V = 0 ˙ b LBasis V Base V = 0 ˙
25 23 24 sseqtrd V LVec Base V = 0 ˙ b LBasis V b 0 ˙
26 sssn b 0 ˙ b = b = 0 ˙
27 25 26 sylib V LVec Base V = 0 ˙ b LBasis V b = b = 0 ˙
28 27 orcomd V LVec Base V = 0 ˙ b LBasis V b = 0 ˙ b =
29 28 ord V LVec Base V = 0 ˙ b LBasis V ¬ b = 0 ˙ b =
30 19 29 mpd V LVec Base V = 0 ˙ b LBasis V b =
31 30 20 eqeltrrd V LVec Base V = 0 ˙ b LBasis V LBasis V
32 8 31 exlimddv V LVec Base V = 0 ˙ LBasis V
33 4 dimval V LVec LBasis V dim V =
34 3 32 33 syl2anc V LVec Base V = 0 ˙ dim V =
35 hash0 = 0
36 34 35 syl6eq V LVec Base V = 0 ˙ dim V = 0
37 2 36 impbida V LVec dim V = 0 Base V = 0 ˙