Metamath Proof Explorer


Theorem lmisfree

Description: A module has a basis iff it is isomorphic to a free module. In settings where isomorphic objects are not distinguished, it is common to define "free module" as any module with a basis; thus for instance lbsex might be described as "every vector space is free". (Contributed by Stefan O'Rear, 26-Feb-2015)

Ref Expression
Hypotheses lmisfree.j J=LBasisW
lmisfree.f F=ScalarW
Assertion lmisfree WLModJkW𝑚FfreeLModk

Proof

Step Hyp Ref Expression
1 lmisfree.j J=LBasisW
2 lmisfree.f F=ScalarW
3 n0 JjjJ
4 vex jV
5 4 enref jj
6 2 1 lbslcic WLModjJjjW𝑚FfreeLModj
7 5 6 mp3an3 WLModjJW𝑚FfreeLModj
8 oveq2 k=jFfreeLModk=FfreeLModj
9 8 breq2d k=jW𝑚FfreeLModkW𝑚FfreeLModj
10 4 9 spcev W𝑚FfreeLModjkW𝑚FfreeLModk
11 7 10 syl WLModjJkW𝑚FfreeLModk
12 11 ex WLModjJkW𝑚FfreeLModk
13 12 exlimdv WLModjjJkW𝑚FfreeLModk
14 3 13 biimtrid WLModJkW𝑚FfreeLModk
15 lmicsym W𝑚FfreeLModkFfreeLModk𝑚W
16 lmiclcl W𝑚FfreeLModkWLMod
17 2 lmodring WLModFRing
18 vex kV
19 eqid FfreeLModk=FfreeLModk
20 eqid FunitVeck=FunitVeck
21 eqid LBasisFfreeLModk=LBasisFfreeLModk
22 19 20 21 frlmlbs FRingkVranFunitVeckLBasisFfreeLModk
23 17 18 22 sylancl WLModranFunitVeckLBasisFfreeLModk
24 23 ne0d WLModLBasisFfreeLModk
25 16 24 syl W𝑚FfreeLModkLBasisFfreeLModk
26 21 1 lmiclbs FfreeLModk𝑚WLBasisFfreeLModkJ
27 15 25 26 sylc W𝑚FfreeLModkJ
28 27 exlimiv kW𝑚FfreeLModkJ
29 14 28 impbid1 WLModJkW𝑚FfreeLModk