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=LHypK
hdmap14lem1a.u U=DVecHKW
hdmap14lem1a.v V=BaseU
hdmap14lem1a.t ·˙=U
hdmap14lem1a.r R=ScalarU
hdmap14lem1a.b B=BaseR
hdmap14lem1a.c C=LCDualKW
hdmap14lem2a.e ˙=C
hdmap14lem1a.l L=LSpanC
hdmap14lem2a.p P=ScalarC
hdmap14lem2a.a A=BaseP
hdmap14lem1a.s S=HDMapKW
hdmap14lem1a.k φKHLWH
hdmap14lem3a.x φXV
hdmap14lem1a.f φFB
Assertion hdmap14lem2a φgASF·˙X=g˙SX

Proof

Step Hyp Ref Expression
1 hdmap14lem1a.h H=LHypK
2 hdmap14lem1a.u U=DVecHKW
3 hdmap14lem1a.v V=BaseU
4 hdmap14lem1a.t ·˙=U
5 hdmap14lem1a.r R=ScalarU
6 hdmap14lem1a.b B=BaseR
7 hdmap14lem1a.c C=LCDualKW
8 hdmap14lem2a.e ˙=C
9 hdmap14lem1a.l L=LSpanC
10 hdmap14lem2a.p P=ScalarC
11 hdmap14lem2a.a A=BaseP
12 hdmap14lem1a.s S=HDMapKW
13 hdmap14lem1a.k φKHLWH
14 hdmap14lem3a.x φXV
15 hdmap14lem1a.f φFB
16 fvoveq1 F=0RSF·˙X=S0R·˙X
17 16 eqeq1d F=0RSF·˙X=g˙SXS0R·˙X=g˙SX
18 17 rexbidv F=0RgASF·˙X=g˙SXgAS0R·˙X=g˙SX
19 difss A0PA
20 13 adantr φF0RKHLWH
21 14 adantr φF0RXV
22 15 adantr φF0RFB
23 eqid 0R=0R
24 simpr φF0RF0R
25 1 2 3 4 5 6 7 8 9 10 11 12 20 21 22 23 24 hdmap14lem1a φF0RLSX=LSF·˙X
26 25 eqcomd φF0RLSF·˙X=LSX
27 eqid BaseC=BaseC
28 eqid 0P=0P
29 1 7 13 lcdlvec φCLVec
30 1 2 13 dvhlmod φULMod
31 3 5 4 6 lmodvscl ULModFBXVF·˙XV
32 30 15 14 31 syl3anc φF·˙XV
33 1 2 3 7 27 12 13 32 hdmapcl φSF·˙XBaseC
34 1 2 3 7 27 12 13 14 hdmapcl φSXBaseC
35 27 10 11 28 8 9 29 33 34 lspsneq φLSF·˙X=LSXgA0PSF·˙X=g˙SX
36 35 adantr φF0RLSF·˙X=LSXgA0PSF·˙X=g˙SX
37 26 36 mpbid φF0RgA0PSF·˙X=g˙SX
38 ssrexv A0PAgA0PSF·˙X=g˙SXgASF·˙X=g˙SX
39 19 37 38 mpsyl φF0RgASF·˙X=g˙SX
40 1 7 13 lcdlmod φCLMod
41 10 11 28 lmod0cl CLMod0PA
42 40 41 syl φ0PA
43 eqid 0U=0U
44 eqid 0C=0C
45 1 2 43 7 44 12 13 hdmapval0 φS0U=0C
46 3 5 4 23 43 lmod0vs ULModXV0R·˙X=0U
47 30 14 46 syl2anc φ0R·˙X=0U
48 47 fveq2d φS0R·˙X=S0U
49 27 10 8 28 44 lmod0vs CLModSXBaseC0P˙SX=0C
50 40 34 49 syl2anc φ0P˙SX=0C
51 45 48 50 3eqtr4d φS0R·˙X=0P˙SX
52 oveq1 g=0Pg˙SX=0P˙SX
53 52 rspceeqv 0PAS0R·˙X=0P˙SXgAS0R·˙X=g˙SX
54 42 51 53 syl2anc φgAS0R·˙X=g˙SX
55 18 39 54 pm2.61ne φgASF·˙X=g˙SX