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