Theorem clatlubcl2

Description: Any subset of the base set has an LUB in a complete lattice. (Contributed by NM, 13-Sep-2018)

	Ref	Expression
Hypotheses	clatlubcl.b	|- B = ( Base ` K )
	clatlubcl.u	|- U = ( lub ` K )
Assertion	clatlubcl2	|- ( ( K e. CLat /\ S C_ B ) -> S e. dom U )

Proof

Step	Hyp	Ref	Expression
1		clatlubcl.b	|- B = ( Base ` K )
2		clatlubcl.u	|- U = ( lub ` K )
3	1	fvexi	|- B e. _V
4	3	elpw2	|- ( S e. ~P B <-> S C_ B )
5	4	biimpri	|- ( S C_ B -> S e. ~P B )
6	5	adantl	|- ( ( K e. CLat /\ S C_ B ) -> S e. ~P B )
7		eqid	|- ( glb ` K ) = ( glb ` K )
8	1 2 7	isclat	|- ( K e. CLat <-> ( K e. Poset /\ ( dom U = ~P B /\ dom ( glb ` K ) = ~P B ) ) )
9		simprl	|- ( ( K e. Poset /\ ( dom U = ~P B /\ dom ( glb ` K ) = ~P B ) ) -> dom U = ~P B )
10	8 9	sylbi	|- ( K e. CLat -> dom U = ~P B )
11	10	adantr	|- ( ( K e. CLat /\ S C_ B ) -> dom U = ~P B )
12	6 11	eleqtrrd	|- ( ( K e. CLat /\ S C_ B ) -> S e. dom U )