let L be Lattice; :: thesis: for S being non empty Subset of L holds
( S is Filter of L iff for p, q being Element of L holds
( ( p in S & q in S ) iff p "/\" q in S ) )
let S be non empty Subset of L; :: thesis: ( S is Filter of L iff for p, q being Element of L holds
( ( p in S & q in S ) iff p "/\" q in S ) )
thus ( S is Filter of L implies for p, q being Element of L holds
( ( p in S & q in S ) iff p "/\" q in S ) ) :: thesis: ( ( for p, q being Element of L holds
( ( p in S & q in S ) iff p "/\" q in S ) ) implies S is Filter of L )

proof

assume A1: S is Filter of L ; :: thesis: for p, q being Element of L holds
( ( p in S & q in S ) iff p "/\" q in S )
let p, q be Element of L; :: thesis: ( ( p in S & q in S ) iff p "/\" q in S )
thus ( p in S & q in S implies p "/\" q in S ) by A1, LATTICES:def 24; :: thesis: ( p "/\" q in S implies ( p in S & q in S ) )
assume A2: p "/\" q in S ; :: thesis: ( p in S & q in S )
( p "/\" q [= p & p "/\" q [= q ) by LATTICES:6;
hence ( p in S & q in S ) by A1, A2, LATTICES:def 23; :: thesis: verum

end;

assume A3: for p, q being Element of L holds
( ( p in S & q in S ) iff p "/\" q in S ) ; :: thesis: S is Filter of L
S is final

proof

let p, q be Element of L; :: according to LATTICES:def 23 :: thesis: ( not p [= q or not p in S or q in S )
assume that
A4: p [= q and
A5: p in S ; :: thesis: q in S
p "/\" q = p by A4, LATTICES:4;
hence q in S by A3, A5; :: thesis: verum

end;

hence S is Filter of L by A3, LATTICES:def 24; :: thesis: verum