Metamath Proof Explorer


Theorem islindf5

Description: A family is independent iff the linear combinations homomorphism is injective. (Contributed by Stefan O'Rear, 28-Feb-2015)

Ref Expression
Hypotheses islindf5.f F = R freeLMod I
islindf5.b B = Base F
islindf5.c C = Base T
islindf5.v · ˙ = T
islindf5.e E = x B T x · ˙ f A
islindf5.t φ T LMod
islindf5.i φ I X
islindf5.r φ R = Scalar T
islindf5.a φ A : I C
Assertion islindf5 φ A LIndF T E : B 1-1 C

Proof

Step Hyp Ref Expression
1 islindf5.f F = R freeLMod I
2 islindf5.b B = Base F
3 islindf5.c C = Base T
4 islindf5.v · ˙ = T
5 islindf5.e E = x B T x · ˙ f A
6 islindf5.t φ T LMod
7 islindf5.i φ I X
8 islindf5.r φ R = Scalar T
9 islindf5.a φ A : I C
10 eqid Scalar T = Scalar T
11 eqid 0 T = 0 T
12 eqid 0 Scalar T = 0 Scalar T
13 eqid Base Scalar T freeLMod I = Base Scalar T freeLMod I
14 3 10 4 11 12 13 islindf4 T LMod I X A : I C A LIndF T y Base Scalar T freeLMod I T y · ˙ f A = 0 T y = I × 0 Scalar T
15 6 7 9 14 syl3anc φ A LIndF T y Base Scalar T freeLMod I T y · ˙ f A = 0 T y = I × 0 Scalar T
16 oveq1 x = y x · ˙ f A = y · ˙ f A
17 16 oveq2d x = y T x · ˙ f A = T y · ˙ f A
18 ovex T y · ˙ f A V
19 17 5 18 fvmpt y B E y = T y · ˙ f A
20 19 adantl φ y B E y = T y · ˙ f A
21 20 eqeq1d φ y B E y = 0 T T y · ˙ f A = 0 T
22 10 lmodring T LMod Scalar T Ring
23 6 22 syl φ Scalar T Ring
24 8 23 eqeltrd φ R Ring
25 eqid 0 R = 0 R
26 1 25 frlm0 R Ring I X I × 0 R = 0 F
27 24 7 26 syl2anc φ I × 0 R = 0 F
28 8 fveq2d φ 0 R = 0 Scalar T
29 28 sneqd φ 0 R = 0 Scalar T
30 29 xpeq2d φ I × 0 R = I × 0 Scalar T
31 27 30 eqtr3d φ 0 F = I × 0 Scalar T
32 31 adantr φ y B 0 F = I × 0 Scalar T
33 32 eqeq2d φ y B y = 0 F y = I × 0 Scalar T
34 21 33 imbi12d φ y B E y = 0 T y = 0 F T y · ˙ f A = 0 T y = I × 0 Scalar T
35 34 ralbidva φ y B E y = 0 T y = 0 F y B T y · ˙ f A = 0 T y = I × 0 Scalar T
36 8 eqcomd φ Scalar T = R
37 36 oveq1d φ Scalar T freeLMod I = R freeLMod I
38 37 1 eqtr4di φ Scalar T freeLMod I = F
39 38 fveq2d φ Base Scalar T freeLMod I = Base F
40 39 2 eqtr4di φ Base Scalar T freeLMod I = B
41 40 raleqdv φ y Base Scalar T freeLMod I T y · ˙ f A = 0 T y = I × 0 Scalar T y B T y · ˙ f A = 0 T y = I × 0 Scalar T
42 35 41 bitr4d φ y B E y = 0 T y = 0 F y Base Scalar T freeLMod I T y · ˙ f A = 0 T y = I × 0 Scalar T
43 15 42 bitr4d φ A LIndF T y B E y = 0 T y = 0 F
44 1 2 3 4 5 6 7 8 9 frlmup1 φ E F LMHom T
45 lmghm E F LMHom T E F GrpHom T
46 eqid 0 F = 0 F
47 2 3 46 11 ghmf1 E F GrpHom T E : B 1-1 C y B E y = 0 T y = 0 F
48 44 45 47 3syl φ E : B 1-1 C y B E y = 0 T y = 0 F
49 43 48 bitr4d φ A LIndF T E : B 1-1 C