Metamath Proof Explorer


Theorem lmicsym

Description: Module isomorphism is symmetric. (Contributed by Stefan O'Rear, 26-Feb-2015)

Ref Expression
Assertion lmicsym
|- ( R ~=m S -> S ~=m R )

Proof

Step Hyp Ref Expression
1 brlmic
 |-  ( R ~=m S <-> ( R LMIso S ) =/= (/) )
2 n0
 |-  ( ( R LMIso S ) =/= (/) <-> E. f f e. ( R LMIso S ) )
3 lmimcnv
 |-  ( f e. ( R LMIso S ) -> `' f e. ( S LMIso R ) )
4 brlmici
 |-  ( `' f e. ( S LMIso R ) -> S ~=m R )
5 3 4 syl
 |-  ( f e. ( R LMIso S ) -> S ~=m R )
6 5 exlimiv
 |-  ( E. f f e. ( R LMIso S ) -> S ~=m R )
7 2 6 sylbi
 |-  ( ( R LMIso S ) =/= (/) -> S ~=m R )
8 1 7 sylbi
 |-  ( R ~=m S -> S ~=m R )