Metamath Proof Explorer


Theorem dvhb1dimN

Description: Two expressions for the 1-dimensional subspaces of vector space H, in the isomorphism B case where the 2nd vector component is zero. (Contributed by NM, 23-Feb-2014) (New usage is discouraged.)

Ref Expression
Hypotheses dvhb1dim.l ˙ = K
dvhb1dim.h H = LHyp K
dvhb1dim.t T = LTrn K W
dvhb1dim.r R = trL K W
dvhb1dim.e E = TEndo K W
dvhb1dim.o 0 ˙ = h T I B
Assertion dvhb1dimN K HL W H F T g T × E | s E g = s F 0 ˙ = g T × E | R 1 st g ˙ R F 2 nd g = 0 ˙

Proof

Step Hyp Ref Expression
1 dvhb1dim.l ˙ = K
2 dvhb1dim.h H = LHyp K
3 dvhb1dim.t T = LTrn K W
4 dvhb1dim.r R = trL K W
5 dvhb1dim.e E = TEndo K W
6 dvhb1dim.o 0 ˙ = h T I B
7 eqop g T × E g = s F 0 ˙ 1 st g = s F 2 nd g = 0 ˙
8 7 adantl K HL W H F T g T × E g = s F 0 ˙ 1 st g = s F 2 nd g = 0 ˙
9 8 rexbidv K HL W H F T g T × E s E g = s F 0 ˙ s E 1 st g = s F 2 nd g = 0 ˙
10 r19.41v s E 1 st g = s F 2 nd g = 0 ˙ s E 1 st g = s F 2 nd g = 0 ˙
11 fvex 1 st g V
12 eqeq1 f = 1 st g f = s F 1 st g = s F
13 12 rexbidv f = 1 st g s E f = s F s E 1 st g = s F
14 11 13 elab 1 st g f | s E f = s F s E 1 st g = s F
15 1 2 3 4 5 dva1dim K HL W H F T f | s E f = s F = f T | R f ˙ R F
16 15 adantr K HL W H F T g T × E f | s E f = s F = f T | R f ˙ R F
17 16 eleq2d K HL W H F T g T × E 1 st g f | s E f = s F 1 st g f T | R f ˙ R F
18 14 17 bitr3id K HL W H F T g T × E s E 1 st g = s F 1 st g f T | R f ˙ R F
19 xp1st g T × E 1 st g T
20 19 adantl K HL W H F T g T × E 1 st g T
21 fveq2 f = 1 st g R f = R 1 st g
22 21 breq1d f = 1 st g R f ˙ R F R 1 st g ˙ R F
23 22 elrab3 1 st g T 1 st g f T | R f ˙ R F R 1 st g ˙ R F
24 20 23 syl K HL W H F T g T × E 1 st g f T | R f ˙ R F R 1 st g ˙ R F
25 18 24 bitrd K HL W H F T g T × E s E 1 st g = s F R 1 st g ˙ R F
26 25 anbi1d K HL W H F T g T × E s E 1 st g = s F 2 nd g = 0 ˙ R 1 st g ˙ R F 2 nd g = 0 ˙
27 10 26 syl5bb K HL W H F T g T × E s E 1 st g = s F 2 nd g = 0 ˙ R 1 st g ˙ R F 2 nd g = 0 ˙
28 9 27 bitrd K HL W H F T g T × E s E g = s F 0 ˙ R 1 st g ˙ R F 2 nd g = 0 ˙
29 28 rabbidva K HL W H F T g T × E | s E g = s F 0 ˙ = g T × E | R 1 st g ˙ R F 2 nd g = 0 ˙