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