Metamath Proof Explorer


Theorem eqlkr3

Description: Two functionals with the same kernel are equal if they are equal at any nonzero value. (Contributed by NM, 2-Jan-2015)

Ref Expression
Hypotheses eqlkr3.v V = Base W
eqlkr3.s S = Scalar W
eqlkr3.r R = Base S
eqlkr3.o 0 ˙ = 0 S
eqlkr3.f F = LFnl W
eqlkr3.k K = LKer W
eqlkr3.w φ W LVec
eqlkr3.x φ X V
eqlkr3.g φ G F
eqlkr3.h φ H F
eqlkr3.e φ K G = K H
eqlkr3.a φ G X = H X
eqlkr3.n φ G X 0 ˙
Assertion eqlkr3 φ G = H

Proof

Step Hyp Ref Expression
1 eqlkr3.v V = Base W
2 eqlkr3.s S = Scalar W
3 eqlkr3.r R = Base S
4 eqlkr3.o 0 ˙ = 0 S
5 eqlkr3.f F = LFnl W
6 eqlkr3.k K = LKer W
7 eqlkr3.w φ W LVec
8 eqlkr3.x φ X V
9 eqlkr3.g φ G F
10 eqlkr3.h φ H F
11 eqlkr3.e φ K G = K H
12 eqlkr3.a φ G X = H X
13 eqlkr3.n φ G X 0 ˙
14 2 3 1 5 lflf W LVec G F G : V R
15 7 9 14 syl2anc φ G : V R
16 15 ffnd φ G Fn V
17 2 3 1 5 lflf W LVec H F H : V R
18 7 10 17 syl2anc φ H : V R
19 18 ffnd φ H Fn V
20 eqid S = S
21 2 3 20 1 5 6 eqlkr W LVec G F H F K G = K H r R x V H x = G x S r
22 7 9 10 11 21 syl121anc φ r R x V H x = G x S r
23 8 adantr φ r R X V
24 fveq2 x = X H x = H X
25 fveq2 x = X G x = G X
26 25 oveq1d x = X G x S r = G X S r
27 24 26 eqeq12d x = X H x = G x S r H X = G X S r
28 27 rspcv X V x V H x = G x S r H X = G X S r
29 23 28 syl φ r R x V H x = G x S r H X = G X S r
30 12 adantr φ r R G X = H X
31 30 adantr φ r R H X = G X S r G X = H X
32 simpr φ r R H X = G X S r H X = G X S r
33 31 32 eqtr2d φ r R H X = G X S r G X S r = G X
34 33 oveq2d φ r R H X = G X S r inv r S G X S G X S r = inv r S G X S G X
35 2 lvecdrng W LVec S DivRing
36 7 35 syl φ S DivRing
37 36 adantr φ r R S DivRing
38 2 3 1 5 lflcl W LVec G F X V G X R
39 7 9 8 38 syl3anc φ G X R
40 39 adantr φ r R G X R
41 13 adantr φ r R G X 0 ˙
42 eqid 1 S = 1 S
43 eqid inv r S = inv r S
44 3 4 20 42 43 drnginvrl S DivRing G X R G X 0 ˙ inv r S G X S G X = 1 S
45 37 40 41 44 syl3anc φ r R inv r S G X S G X = 1 S
46 45 oveq1d φ r R inv r S G X S G X S r = 1 S S r
47 lveclmod W LVec W LMod
48 7 47 syl φ W LMod
49 2 lmodring W LMod S Ring
50 48 49 syl φ S Ring
51 50 adantr φ r R S Ring
52 3 4 43 drnginvrcl S DivRing G X R G X 0 ˙ inv r S G X R
53 37 40 41 52 syl3anc φ r R inv r S G X R
54 simpr φ r R r R
55 3 20 ringass S Ring inv r S G X R G X R r R inv r S G X S G X S r = inv r S G X S G X S r
56 51 53 40 54 55 syl13anc φ r R inv r S G X S G X S r = inv r S G X S G X S r
57 3 20 42 ringlidm S Ring r R 1 S S r = r
58 51 54 57 syl2anc φ r R 1 S S r = r
59 46 56 58 3eqtr3d φ r R inv r S G X S G X S r = r
60 59 adantr φ r R H X = G X S r inv r S G X S G X S r = r
61 45 adantr φ r R H X = G X S r inv r S G X S G X = 1 S
62 34 60 61 3eqtr3d φ r R H X = G X S r r = 1 S
63 62 ex φ r R H X = G X S r r = 1 S
64 29 63 syld φ r R x V H x = G x S r r = 1 S
65 64 ancrd φ r R x V H x = G x S r r = 1 S x V H x = G x S r
66 65 reximdva φ r R x V H x = G x S r r R r = 1 S x V H x = G x S r
67 22 66 mpd φ r R r = 1 S x V H x = G x S r
68 3 42 ringidcl S Ring 1 S R
69 50 68 syl φ 1 S R
70 oveq2 r = 1 S G x S r = G x S 1 S
71 70 eqeq2d r = 1 S H x = G x S r H x = G x S 1 S
72 71 ralbidv r = 1 S x V H x = G x S r x V H x = G x S 1 S
73 72 ceqsrexv 1 S R r R r = 1 S x V H x = G x S r x V H x = G x S 1 S
74 69 73 syl φ r R r = 1 S x V H x = G x S r x V H x = G x S 1 S
75 67 74 mpbid φ x V H x = G x S 1 S
76 75 r19.21bi φ x V H x = G x S 1 S
77 48 adantr φ x V W LMod
78 77 49 syl φ x V S Ring
79 7 adantr φ x V W LVec
80 9 adantr φ x V G F
81 simpr φ x V x V
82 2 3 1 5 lflcl W LVec G F x V G x R
83 79 80 81 82 syl3anc φ x V G x R
84 3 20 42 ringridm S Ring G x R G x S 1 S = G x
85 78 83 84 syl2anc φ x V G x S 1 S = G x
86 76 85 eqtr2d φ x V G x = H x
87 16 19 86 eqfnfvd φ G = H