Theorem mndid

Description: A monoid has a two-sided identity element. (Contributed by NM, 16-Aug-2011)

	Ref	Expression
Hypotheses	mndcl.b	|- B = ( Base ` G )
	mndcl.p	|- .+ = ( +g ` G )
Assertion	mndid	|- ( G e. Mnd -> E. u e. B A. x e. B ( ( u .+ x ) = x /\ ( x .+ u ) = x ) )

Proof

Step	Hyp	Ref	Expression
1		mndcl.b	|- B = ( Base ` G )
2		mndcl.p	|- .+ = ( +g ` G )
3	1 2	ismnd	|- ( G e. Mnd <-> ( A. x e. B A. y e. B ( ( x .+ y ) e. B /\ A. z e. B ( ( x .+ y ) .+ z ) = ( x .+ ( y .+ z ) ) ) /\ E. u e. B A. x e. B ( ( u .+ x ) = x /\ ( x .+ u ) = x ) ) )
4	3	simprbi	|- ( G e. Mnd -> E. u e. B A. x e. B ( ( u .+ x ) = x /\ ( x .+ u ) = x ) )