Metamath Proof Explorer


Theorem linc0scn0

Description: If a set contains the zero element of a module, there is a linear combination being 0 where not all scalars are 0. (Contributed by AV, 13-Apr-2019)

Ref Expression
Hypotheses linc0scn0.b B = Base M
linc0scn0.s S = Scalar M
linc0scn0.0 0 ˙ = 0 S
linc0scn0.1 1 ˙ = 1 S
linc0scn0.z Z = 0 M
linc0scn0.f F = x V if x = Z 1 ˙ 0 ˙
Assertion linc0scn0 M LMod V 𝒫 B F linC M V = Z

Proof

Step Hyp Ref Expression
1 linc0scn0.b B = Base M
2 linc0scn0.s S = Scalar M
3 linc0scn0.0 0 ˙ = 0 S
4 linc0scn0.1 1 ˙ = 1 S
5 linc0scn0.z Z = 0 M
6 linc0scn0.f F = x V if x = Z 1 ˙ 0 ˙
7 simpl M LMod V 𝒫 B M LMod
8 2 lmodring M LMod S Ring
9 2 eqcomi Scalar M = S
10 9 fveq2i Base Scalar M = Base S
11 10 4 ringidcl S Ring 1 ˙ Base Scalar M
12 10 3 ring0cl S Ring 0 ˙ Base Scalar M
13 11 12 jca S Ring 1 ˙ Base Scalar M 0 ˙ Base Scalar M
14 8 13 syl M LMod 1 ˙ Base Scalar M 0 ˙ Base Scalar M
15 14 ad2antrr M LMod V 𝒫 B x V 1 ˙ Base Scalar M 0 ˙ Base Scalar M
16 ifcl 1 ˙ Base Scalar M 0 ˙ Base Scalar M if x = Z 1 ˙ 0 ˙ Base Scalar M
17 15 16 syl M LMod V 𝒫 B x V if x = Z 1 ˙ 0 ˙ Base Scalar M
18 17 6 fmptd M LMod V 𝒫 B F : V Base Scalar M
19 fvex Base Scalar M V
20 19 a1i M LMod Base Scalar M V
21 elmapg Base Scalar M V V 𝒫 B F Base Scalar M V F : V Base Scalar M
22 20 21 sylan M LMod V 𝒫 B F Base Scalar M V F : V Base Scalar M
23 18 22 mpbird M LMod V 𝒫 B F Base Scalar M V
24 1 pweqi 𝒫 B = 𝒫 Base M
25 24 eleq2i V 𝒫 B V 𝒫 Base M
26 25 bilani M LMod V 𝒫 B V 𝒫 Base M
27 lincval M LMod F Base Scalar M V V 𝒫 Base M F linC M V = M v V F v M v
28 7 23 26 27 syl3anc M LMod V 𝒫 B F linC M V = M v V F v M v
29 simpr M LMod V 𝒫 B v V v V
30 4 fvexi 1 ˙ V
31 3 fvexi 0 ˙ V
32 30 31 ifex if v = Z 1 ˙ 0 ˙ V
33 eqeq1 x = v x = Z v = Z
34 33 ifbid x = v if x = Z 1 ˙ 0 ˙ = if v = Z 1 ˙ 0 ˙
35 34 6 fvmptg v V if v = Z 1 ˙ 0 ˙ V F v = if v = Z 1 ˙ 0 ˙
36 29 32 35 sylancl M LMod V 𝒫 B v V F v = if v = Z 1 ˙ 0 ˙
37 36 oveq1d M LMod V 𝒫 B v V F v M v = if v = Z 1 ˙ 0 ˙ M v
38 ovif if v = Z 1 ˙ 0 ˙ M v = if v = Z 1 ˙ M v 0 ˙ M v
39 38 a1i M LMod V 𝒫 B v V if v = Z 1 ˙ 0 ˙ M v = if v = Z 1 ˙ M v 0 ˙ M v
40 oveq2 v = Z 1 ˙ M v = 1 ˙ M Z
41 40 adantl M LMod V 𝒫 B v V v = Z 1 ˙ M v = 1 ˙ M Z
42 eqid Base S = Base S
43 2 42 4 lmod1cl M LMod 1 ˙ Base S
44 43 ancli M LMod M LMod 1 ˙ Base S
45 44 adantr M LMod V 𝒫 B M LMod 1 ˙ Base S
46 45 ad2antrr M LMod V 𝒫 B v V v = Z M LMod 1 ˙ Base S
47 eqid M = M
48 2 47 42 5 lmodvs0 M LMod 1 ˙ Base S 1 ˙ M Z = Z
49 46 48 syl M LMod V 𝒫 B v V v = Z 1 ˙ M Z = Z
50 41 49 eqtrd M LMod V 𝒫 B v V v = Z 1 ˙ M v = Z
51 7 adantr M LMod V 𝒫 B v V M LMod
52 elelpwi v V V 𝒫 B v B
53 52 expcom V 𝒫 B v V v B
54 53 adantl M LMod V 𝒫 B v V v B
55 54 imp M LMod V 𝒫 B v V v B
56 1 2 47 3 5 lmod0vs M LMod v B 0 ˙ M v = Z
57 51 55 56 syl2anc M LMod V 𝒫 B v V 0 ˙ M v = Z
58 57 adantr M LMod V 𝒫 B v V ¬ v = Z 0 ˙ M v = Z
59 50 58 ifeqda M LMod V 𝒫 B v V if v = Z 1 ˙ M v 0 ˙ M v = Z
60 37 39 59 3eqtrd M LMod V 𝒫 B v V F v M v = Z
61 60 mpteq2dva M LMod V 𝒫 B v V F v M v = v V Z
62 61 oveq2d M LMod V 𝒫 B M v V F v M v = M v V Z
63 lmodgrp M LMod M Grp
64 63 grpmndd M LMod M Mnd
65 5 gsumz M Mnd V 𝒫 B M v V Z = Z
66 64 65 sylan M LMod V 𝒫 B M v V Z = Z
67 28 62 66 3eqtrd M LMod V 𝒫 B F linC M V = Z