Metamath Proof Explorer


Theorem hdmap14lem2a

Description: Prior to part 14 in Baer p. 49, line 25. TODO: fix to include F = .0. so it can be used in hdmap14lem10 . (Contributed by NM, 31-May-2015)

Ref Expression
Hypotheses hdmap14lem1a.h H = LHyp K
hdmap14lem1a.u U = DVecH K W
hdmap14lem1a.v V = Base U
hdmap14lem1a.t · ˙ = U
hdmap14lem1a.r R = Scalar U
hdmap14lem1a.b B = Base R
hdmap14lem1a.c C = LCDual K W
hdmap14lem2a.e ˙ = C
hdmap14lem1a.l L = LSpan C
hdmap14lem2a.p P = Scalar C
hdmap14lem2a.a A = Base P
hdmap14lem1a.s S = HDMap K W
hdmap14lem1a.k φ K HL W H
hdmap14lem3a.x φ X V
hdmap14lem1a.f φ F B
Assertion hdmap14lem2a φ g A S F · ˙ X = g ˙ S X

Proof

Step Hyp Ref Expression
1 hdmap14lem1a.h H = LHyp K
2 hdmap14lem1a.u U = DVecH K W
3 hdmap14lem1a.v V = Base U
4 hdmap14lem1a.t · ˙ = U
5 hdmap14lem1a.r R = Scalar U
6 hdmap14lem1a.b B = Base R
7 hdmap14lem1a.c C = LCDual K W
8 hdmap14lem2a.e ˙ = C
9 hdmap14lem1a.l L = LSpan C
10 hdmap14lem2a.p P = Scalar C
11 hdmap14lem2a.a A = Base P
12 hdmap14lem1a.s S = HDMap K W
13 hdmap14lem1a.k φ K HL W H
14 hdmap14lem3a.x φ X V
15 hdmap14lem1a.f φ F B
16 fvoveq1 F = 0 R S F · ˙ X = S 0 R · ˙ X
17 16 eqeq1d F = 0 R S F · ˙ X = g ˙ S X S 0 R · ˙ X = g ˙ S X
18 17 rexbidv F = 0 R g A S F · ˙ X = g ˙ S X g A S 0 R · ˙ X = g ˙ S X
19 difss A 0 P A
20 13 adantr φ F 0 R K HL W H
21 14 adantr φ F 0 R X V
22 15 adantr φ F 0 R F B
23 eqid 0 R = 0 R
24 simpr φ F 0 R F 0 R
25 1 2 3 4 5 6 7 8 9 10 11 12 20 21 22 23 24 hdmap14lem1a φ F 0 R L S X = L S F · ˙ X
26 25 eqcomd φ F 0 R L S F · ˙ X = L S X
27 eqid Base C = Base C
28 eqid 0 P = 0 P
29 1 7 13 lcdlvec φ C LVec
30 1 2 13 dvhlmod φ U LMod
31 3 5 4 6 lmodvscl U LMod F B X V F · ˙ X V
32 30 15 14 31 syl3anc φ F · ˙ X V
33 1 2 3 7 27 12 13 32 hdmapcl φ S F · ˙ X Base C
34 1 2 3 7 27 12 13 14 hdmapcl φ S X Base C
35 27 10 11 28 8 9 29 33 34 lspsneq φ L S F · ˙ X = L S X g A 0 P S F · ˙ X = g ˙ S X
36 35 adantr φ F 0 R L S F · ˙ X = L S X g A 0 P S F · ˙ X = g ˙ S X
37 26 36 mpbid φ F 0 R g A 0 P S F · ˙ X = g ˙ S X
38 ssrexv A 0 P A g A 0 P S F · ˙ X = g ˙ S X g A S F · ˙ X = g ˙ S X
39 19 37 38 mpsyl φ F 0 R g A S F · ˙ X = g ˙ S X
40 1 7 13 lcdlmod φ C LMod
41 10 11 28 lmod0cl C LMod 0 P A
42 40 41 syl φ 0 P A
43 eqid 0 U = 0 U
44 eqid 0 C = 0 C
45 1 2 43 7 44 12 13 hdmapval0 φ S 0 U = 0 C
46 3 5 4 23 43 lmod0vs U LMod X V 0 R · ˙ X = 0 U
47 30 14 46 syl2anc φ 0 R · ˙ X = 0 U
48 47 fveq2d φ S 0 R · ˙ X = S 0 U
49 27 10 8 28 44 lmod0vs C LMod S X Base C 0 P ˙ S X = 0 C
50 40 34 49 syl2anc φ 0 P ˙ S X = 0 C
51 45 48 50 3eqtr4d φ S 0 R · ˙ X = 0 P ˙ S X
52 oveq1 g = 0 P g ˙ S X = 0 P ˙ S X
53 52 rspceeqv 0 P A S 0 R · ˙ X = 0 P ˙ S X g A S 0 R · ˙ X = g ˙ S X
54 42 51 53 syl2anc φ g A S 0 R · ˙ X = g ˙ S X
55 18 39 54 pm2.61ne φ g A S F · ˙ X = g ˙ S X