reserve L,L1,L2 for Lattice,
  F1,F2 for Filter of L,
  p,q,r,s for Element of L,
  p1,q1,r1,s1 for Element of L1,
  p2,q2,r2,s2 for Element of L2,
  X,x,x1,x2,y,y1,y2 for set,
  D,D1,D2 for non empty set,
  R for Relation,
  RD for Equivalence_Relation of D,
  a,b,d for Element of D,
  a1,b1,c1 for Element of D1,
  a2,b2,c2 for Element of D2,
  B for B_Lattice,
  FB for Filter of B,
  I for I_Lattice,
  FI for Filter of I ,
  i,i1,i2,j,j1,j2,k for Element of I,
  f1,g1 for BinOp of D1,
  f2,g2 for BinOp of D2;
reserve F,G for BinOp of D,RD;

theorem Th42:
  L1 is 0_Lattice & L2 is 0_Lattice implies Bottom [:L1,L2:] = [
  Bottom L1, Bottom L2]
proof
  assume that
A1: L1 is 0_Lattice and
A2: L2 is 0_Lattice;
A3: now
    let a be Element of [:L1,L2:];
    consider p1,p2 such that
A4: a = [p1,p2] by DOMAIN_1:1;
    thus [Bottom L1,Bottom L2]"/\"a = [Bottom L1"/\"p1,Bottom L2"/\"p2] by A4
,Th21
      .= [Bottom L1,Bottom L2"/\" p2] by A1
      .= [Bottom L1,Bottom L2] by A2;
    hence a"/\"[Bottom L1,Bottom L2]=[Bottom L1,Bottom L2];
  end;
  [:L1,L2:] is lower-bounded by A1,A2,Th39;
  hence thesis by A3,LATTICES:def 16;
end;
