Metamath Proof Explorer


Theorem lvecisfrlm

Description: Every vector space is isomorphic to a free module. (Contributed by AV, 7-Mar-2019)

Ref Expression
Hypothesis lvecisfrlm.f ⊢ F = Scalar ⁡ W
Assertion lvecisfrlm ⊢ W ∈ LVec → ∃ k W ≃ 𝑚 F freeLMod k

Proof

Step Hyp Ref Expression
1 lvecisfrlm.f ⊢ F = Scalar ⁡ W
2 eqid ⊢ LBasis ⁡ W = LBasis ⁡ W
3 2 lbsex ⊢ W ∈ LVec → LBasis ⁡ W ≠ ∅
4 lveclmod ⊢ W ∈ LVec → W ∈ LMod
5 2 1 lmisfree ⊢ W ∈ LMod → LBasis ⁡ W ≠ ∅ ↔ ∃ k W ≃ 𝑚 F freeLMod k
6 4 5 syl ⊢ W ∈ LVec → LBasis ⁡ W ≠ ∅ ↔ ∃ k W ≃ 𝑚 F freeLMod k
7 3 6 mpbid ⊢ W ∈ LVec → ∃ k W ≃ 𝑚 F freeLMod k