Theorem lmiclcl

Description: Isomorphism implies the left side is a module. (Contributed by Stefan O'Rear, 25-Jan-2015)

	Ref	Expression
Assertion	lmiclcl	|- ( R ~=m S -> R e. LMod )

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	1 2	bitri	|- ( R ~=m S <-> E. f f e. ( R LMIso S ) )
4		lmimlmhm	|- ( f e. ( R LMIso S ) -> f e. ( R LMHom S ) )
5		lmhmlmod1	|- ( f e. ( R LMHom S ) -> R e. LMod )
6	4 5	syl	|- ( f e. ( R LMIso S ) -> R e. LMod )
7	6	exlimiv	|- ( E. f f e. ( R LMIso S ) -> R e. LMod )
8	3 7	sylbi	|- ( R ~=m S -> R e. LMod )