 reserve L for AD_Lattice;
 reserve x,y,z for Element of L;
 reserve L for GAD_Lattice;
 reserve x,y,z for Element of L;
reserve L for with_zero GAD_Lattice,
        x,y for Element of L;

theorem :: Corollary 3.12
  for a being Element of L, X being set st
    X = the set of all x "/\" a where x is Element of L
  holds
    X = {x where x is Element of L: x [= a} & X is ClosedSubset of L
  proof
    let a be Element of L;
    let X be set;
    assume
B1: X = the set of all x "/\" a where x is Element of L;
B2: for o being object holds
    o in the set of all x "/\" a where x is Element of L iff
    o in {x where x is Element of L: x [= a}
    proof
      let o be object;
      hereby
        assume o in the set of all x "/\" a where x is Element of L;
        then consider x be Element of L such that
A1:     o = x "/\" a;
        set y = x "/\" a;
        L is meet-absorbing;
        then y [= a;
        hence o in {y where y is Element of L: y [= a} by A1;
      end;
      assume o in {x where x is Element of L: x [= a};
      then consider x be Element of L such that
A1:   o = x & x [= a;
      x "/\" a = x by A1,LATTICES:def 9;
      hence o in the set of all x "/\" a where x is Element of L by A1;
    end;
    now
      let o be object;
      assume o in X;
      then consider x be Element of L such that
B3:   o = x "/\" a by B1;
      thus o in the carrier of L by B3;
    end;
    then reconsider S = X as Subset of L by TARSKI:def 3;
    for p,q being Element of L st p in S & q in S holds p "/\" q in S
    proof
      let p,q be Element of L;
      assume p in S;
      then consider x be Element of L such that
B5:   p = x "/\" a by B1;
      assume q in S;
      then consider y be Element of L such that
B6:   q = y "/\" a by B1;
      p "/\" q = x "/\" (a "/\" (y "/\" a)) by B5,B6,LATTICES:def 7
      .= x "/\" ((a "/\" y) "/\" a) by LATTICES:def 7
      .= (x "/\" (a "/\" y)) "/\" a by LATTICES:def 7;
      hence p "/\" q in S by B1;
    end;
    then
B4: S is meet-closed;
    for p,q being Element of L st p in S & q in S holds p "\/" q in S
    proof
      let p,q be Element of L;
      assume p in S;
      then consider x be Element of L such that
B5:   p = x "/\" a by B1;
      assume q in S;
      then consider y be Element of L such that
B6:   q = y "/\" a by B1;
      p "\/" q = ((x "/\" a) "\/" y) "/\" ((x "/\" a) "\/" a) by B5,B6,DefLDS
      .= ((x "/\" a) "\/" y) "/\" a by LATTICES:def 8;
      hence p "\/" q in S by B1;
    end;
    then S is join-closed;
    hence thesis by B1,B2,B4,TARSKI:2;
  end;
