Metamath Proof Explorer


Theorem lcfl7lem

Description: Lemma for lcfl7N . If two functionals G and J are equal, they are determined by the same vector. (Contributed by NM, 4-Jan-2015)

Ref Expression
Hypotheses lcfl7lem.h H=LHypK
lcfl7lem.o ˙=ocHKW
lcfl7lem.u U=DVecHKW
lcfl7lem.v V=BaseU
lcfl7lem.a +˙=+U
lcfl7lem.t ·˙=U
lcfl7lem.s S=ScalarU
lcfl7lem.r R=BaseS
lcfl7lem.z 0˙=0U
lcfl7lem.f F=LFnlU
lcfl7lem.l L=LKerU
lcfl7lem.k φKHLWH
lcfl7lem.g G=vVιkR|w˙Xv=w+˙k·˙X
lcfl7lem.j J=vVιkR|w˙Yv=w+˙k·˙Y
lcfl7lem.x φXV0˙
lcfl7lem.x2 φYV0˙
lcfl7lem.gj φG=J
Assertion lcfl7lem φX=Y

Proof

Step Hyp Ref Expression
1 lcfl7lem.h H=LHypK
2 lcfl7lem.o ˙=ocHKW
3 lcfl7lem.u U=DVecHKW
4 lcfl7lem.v V=BaseU
5 lcfl7lem.a +˙=+U
6 lcfl7lem.t ·˙=U
7 lcfl7lem.s S=ScalarU
8 lcfl7lem.r R=BaseS
9 lcfl7lem.z 0˙=0U
10 lcfl7lem.f F=LFnlU
11 lcfl7lem.l L=LKerU
12 lcfl7lem.k φKHLWH
13 lcfl7lem.g G=vVιkR|w˙Xv=w+˙k·˙X
14 lcfl7lem.j J=vVιkR|w˙Yv=w+˙k·˙Y
15 lcfl7lem.x φXV0˙
16 lcfl7lem.x2 φYV0˙
17 lcfl7lem.gj φG=J
18 1 2 3 4 9 5 6 11 7 8 13 12 15 dochsnkr2cl φX˙LG0˙
19 18 eldifad φX˙LG
20 17 fveq2d φLG=LJ
21 1 2 3 4 9 5 6 11 7 8 14 12 16 dochsnkr2 φLJ=˙Y
22 20 21 eqtrd φLG=˙Y
23 22 fveq2d φ˙LG=˙˙Y
24 eqid LSpanU=LSpanU
25 16 eldifad φYV
26 25 snssd φYV
27 1 3 2 4 24 12 26 dochocsp φ˙LSpanUY=˙Y
28 27 fveq2d φ˙˙LSpanUY=˙˙Y
29 eqid DIsoHKW=DIsoHKW
30 1 3 4 24 29 dihlsprn KHLWHYVLSpanUYranDIsoHKW
31 12 25 30 syl2anc φLSpanUYranDIsoHKW
32 1 29 2 dochoc KHLWHLSpanUYranDIsoHKW˙˙LSpanUY=LSpanUY
33 12 31 32 syl2anc φ˙˙LSpanUY=LSpanUY
34 23 28 33 3eqtr2d φ˙LG=LSpanUY
35 19 34 eleqtrd φXLSpanUY
36 1 3 12 dvhlmod φULMod
37 7 8 4 6 24 lspsnel ULModYVXLSpanUYsRX=s·˙Y
38 36 25 37 syl2anc φXLSpanUYsRX=s·˙Y
39 35 38 mpbid φsRX=s·˙Y
40 simp3 φsRX=s·˙YX=s·˙Y
41 fveq2 X=s·˙YGX=Gs·˙Y
42 41 3ad2ant3 φsRX=s·˙YGX=Gs·˙Y
43 eqid 1S=1S
44 1 2 3 4 5 6 9 7 8 43 12 16 14 dochfl1 φJY=1S
45 17 fveq1d φGY=JY
46 1 2 3 4 5 6 9 7 8 43 12 15 13 dochfl1 φGX=1S
47 44 45 46 3eqtr4rd φGX=GY
48 47 3ad2ant1 φsRX=s·˙YGX=GY
49 36 3ad2ant1 φsRX=s·˙YULMod
50 1 2 3 4 9 5 6 10 7 8 13 12 15 dochflcl φGF
51 50 3ad2ant1 φsRX=s·˙YGF
52 simp2 φsRX=s·˙YsR
53 25 3ad2ant1 φsRX=s·˙YYV
54 eqid S=S
55 7 8 54 4 6 10 lflmul ULModGFsRYVGs·˙Y=sSGY
56 49 51 52 53 55 syl112anc φsRX=s·˙YGs·˙Y=sSGY
57 42 48 56 3eqtr3d φsRX=s·˙YGY=sSGY
58 57 oveq1d φsRX=s·˙YGYSinvrSGY=sSGYSinvrSGY
59 7 lmodring ULModSRing
60 36 59 syl φSRing
61 60 3ad2ant1 φsRX=s·˙YSRing
62 7 8 4 10 lflcl ULModGFYVGYR
63 36 50 25 62 syl3anc φGYR
64 63 3ad2ant1 φsRX=s·˙YGYR
65 1 3 12 dvhlvec φULVec
66 7 lvecdrng ULVecSDivRing
67 65 66 syl φSDivRing
68 45 44 eqtrd φGY=1S
69 eqid 0S=0S
70 69 43 drngunz SDivRing1S0S
71 67 70 syl φ1S0S
72 68 71 eqnetrd φGY0S
73 eqid invrS=invrS
74 8 69 73 drnginvrcl SDivRingGYRGY0SinvrSGYR
75 67 63 72 74 syl3anc φinvrSGYR
76 75 3ad2ant1 φsRX=s·˙YinvrSGYR
77 8 54 ringass SRingsRGYRinvrSGYRsSGYSinvrSGY=sSGYSinvrSGY
78 61 52 64 76 77 syl13anc φsRX=s·˙YsSGYSinvrSGY=sSGYSinvrSGY
79 8 69 54 43 73 drnginvrr SDivRingGYRGY0SGYSinvrSGY=1S
80 67 63 72 79 syl3anc φGYSinvrSGY=1S
81 80 3ad2ant1 φsRX=s·˙YGYSinvrSGY=1S
82 81 oveq2d φsRX=s·˙YsSGYSinvrSGY=sS1S
83 58 78 82 3eqtrrd φsRX=s·˙YsS1S=GYSinvrSGY
84 8 54 43 ringridm SRingsRsS1S=s
85 61 52 84 syl2anc φsRX=s·˙YsS1S=s
86 83 85 81 3eqtr3d φsRX=s·˙Ys=1S
87 oveq1 s=1Ss·˙Y=1S·˙Y
88 4 7 6 43 lmodvs1 ULModYV1S·˙Y=Y
89 36 25 88 syl2anc φ1S·˙Y=Y
90 89 3ad2ant1 φsRX=s·˙Y1S·˙Y=Y
91 87 90 sylan9eqr φsRX=s·˙Ys=1Ss·˙Y=Y
92 86 91 mpdan φsRX=s·˙Ys·˙Y=Y
93 40 92 eqtrd φsRX=s·˙YX=Y
94 93 rexlimdv3a φsRX=s·˙YX=Y
95 39 94 mpd φX=Y