L: a lattice
Q(L) = quotient set of the lattice = {b/a | a in J(L) and b in M(L) and a <= b}




Proposition 1
If S is a suballtice L of finite length then
S = L - cup_A [a,b] for some A \subseteq Q(L).

Proof: 
to show that for all x in L-S, there is some b/a in Q(L) such that x \in [a,b] and [a,b] \subseteq L-S.

If not, 

exists x on L-S: no such b/a exists.  (*)


Let A = {a in J(L) | a <= x}
Let B = {b in M(L) | x <= b}

A is non-empty.
B is no-empty


From (*), for all [a,b] such that a in A and b in B, they have non-empty intersection with S

Let yba be in S \cap [a,b]

   Let z = join_a meet_b yba \in S   


because S is a complete sub-lattice, z \in S.


We will also show that z is in L-S a contradiction.

   x
=		{ definition of A }
  join_{a in A} (a)
= 		{idempotency of meets}
  join_{a in A} meet_{b in B} (a)
<=		{ a <= yba }
  join_{a in A} meet_{b in B} (yba)
<=		{ yba <= b }
  join_{a in A} meet_{b in B} (b)

=
  meet_{b in B} b
=
  x

This implies all must be equal. Therefore, z=x which is in L-S.


end of proof.

More direct proof is as follows.