Metamath Proof Explorer


Theorem lindssnlvec

Description: A singleton not containing the zero element of a vector space is always linearly independent. (Contributed by AV, 16-Apr-2019) (Revised by AV, 28-Apr-2019)

Ref Expression
Assertion lindssnlvec ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M → S linIndS M

Proof

Step Hyp Ref Expression
1 eldifsni ⊢ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M → s ≠ 0 Scalar ⁡ M
2 1 adantl ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M ∧ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M → s ≠ 0 Scalar ⁡ M
3 simpl3 ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M ∧ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M → S ≠ 0 M
4 eqid ⊢ Base M = Base M
5 eqid ⊢ ⋅ M = ⋅ M
6 eqid ⊢ Scalar ⁡ M = Scalar ⁡ M
7 eqid ⊢ Base Scalar ⁡ M = Base Scalar ⁡ M
8 eqid ⊢ 0 Scalar ⁡ M = 0 Scalar ⁡ M
9 eqid ⊢ 0 M = 0 M
10 simpl1 ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M ∧ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M → M ∈ LVec
11 eldifi ⊢ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M → s ∈ Base Scalar ⁡ M
12 11 adantl ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M ∧ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M → s ∈ Base Scalar ⁡ M
13 simpl2 ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M ∧ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M → S ∈ Base M
14 4 5 6 7 8 9 10 12 13 lvecvsn0 ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M ∧ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M → s ⋅ M S ≠ 0 M ↔ s ≠ 0 Scalar ⁡ M ∧ S ≠ 0 M
15 2 3 14 mpbir2and ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M ∧ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M → s ⋅ M S ≠ 0 M
16 15 ralrimiva ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M → ∀ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M s ⋅ M S ≠ 0 M
17 lveclmod ⊢ M ∈ LVec → M ∈ LMod
18 17 anim1i ⊢ M ∈ LVec ∧ S ∈ Base M → M ∈ LMod ∧ S ∈ Base M
19 18 3adant3 ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M → M ∈ LMod ∧ S ∈ Base M
20 4 6 7 8 9 5 snlindsntor ⊢ M ∈ LMod ∧ S ∈ Base M → ∀ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M s ⋅ M S ≠ 0 M ↔ S linIndS M
21 19 20 syl ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M → ∀ s ∈ Base Scalar ⁡ M ∖ 0 Scalar ⁡ M s ⋅ M S ≠ 0 M ↔ S linIndS M
22 16 21 mpbid ⊢ M ∈ LVec ∧ S ∈ Base M ∧ S ≠ 0 M → S linIndS M