Metamath Proof Explorer


Theorem snlindsntorlem

Description: Lemma for snlindsntor . (Contributed by AV, 15-Apr-2019)

Ref Expression
Hypotheses snlindsntor.b ⊢ B = Base M
snlindsntor.r ⊢ R = Scalar ⁡ M
snlindsntor.s ⊢ S = Base R
snlindsntor.0 ⊢ 0 ˙ = 0 R
snlindsntor.z ⊢ Z = 0 M
snlindsntor.t ⊢ · ˙ = ⋅ M
Assertion snlindsntorlem ⊢ M ∈ LMod ∧ X ∈ B → ∀ f ∈ S X f linC ⁡ M X = Z → f ⁡ X = 0 ˙ → ∀ s ∈ S s · ˙ X = Z → s = 0 ˙

Proof

Step Hyp Ref Expression
1 snlindsntor.b ⊢ B = Base M
2 snlindsntor.r ⊢ R = Scalar ⁡ M
3 snlindsntor.s ⊢ S = Base R
4 snlindsntor.0 ⊢ 0 ˙ = 0 R
5 snlindsntor.z ⊢ Z = 0 M
6 snlindsntor.t ⊢ · ˙ = ⋅ M
7 eqidd ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s = X s
8 fsng ⊢ X ∈ B ∧ s ∈ S → X s : X ⟶ s ↔ X s = X s
9 8 adantll ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s : X ⟶ s ↔ X s = X s
10 7 9 mpbird ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s : X ⟶ s
11 snssi ⊢ s ∈ S → s ⊆ S
12 11 adantl ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → s ⊆ S
13 10 12 fssd ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s : X ⟶ S
14 3 fvexi ⊢ S ∈ V
15 snex ⊢ X ∈ V
16 14 15 pm3.2i ⊢ S ∈ V ∧ X ∈ V
17 elmapg ⊢ S ∈ V ∧ X ∈ V → X s ∈ S X ↔ X s : X ⟶ S
18 16 17 mp1i ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s ∈ S X ↔ X s : X ⟶ S
19 13 18 mpbird ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s ∈ S X
20 oveq1 ⊢ f = X s → f linC ⁡ M X = X s linC ⁡ M X
21 20 eqeq1d ⊢ f = X s → f linC ⁡ M X = Z ↔ X s linC ⁡ M X = Z
22 fveq1 ⊢ f = X s → f ⁡ X = X s ⁡ X
23 22 eqeq1d ⊢ f = X s → f ⁡ X = 0 ˙ ↔ X s ⁡ X = 0 ˙
24 21 23 imbi12d ⊢ f = X s → f linC ⁡ M X = Z → f ⁡ X = 0 ˙ ↔ X s linC ⁡ M X = Z → X s ⁡ X = 0 ˙
25 1 2 3 6 lincvalsng ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s linC ⁡ M X = s · ˙ X
26 25 3expa ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s linC ⁡ M X = s · ˙ X
27 26 eqeq1d ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s linC ⁡ M X = Z ↔ s · ˙ X = Z
28 fvsng ⊢ X ∈ B ∧ s ∈ S → X s ⁡ X = s
29 28 adantll ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s ⁡ X = s
30 29 eqeq1d ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s ⁡ X = 0 ˙ ↔ s = 0 ˙
31 27 30 imbi12d ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → X s linC ⁡ M X = Z → X s ⁡ X = 0 ˙ ↔ s · ˙ X = Z → s = 0 ˙
32 24 31 sylan9bbr ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S ∧ f = X s → f linC ⁡ M X = Z → f ⁡ X = 0 ˙ ↔ s · ˙ X = Z → s = 0 ˙
33 19 32 rspcdv ⊢ M ∈ LMod ∧ X ∈ B ∧ s ∈ S → ∀ f ∈ S X f linC ⁡ M X = Z → f ⁡ X = 0 ˙ → s · ˙ X = Z → s = 0 ˙
34 33 ralrimdva ⊢ M ∈ LMod ∧ X ∈ B → ∀ f ∈ S X f linC ⁡ M X = Z → f ⁡ X = 0 ˙ → ∀ s ∈ S s · ˙ X = Z → s = 0 ˙