This theorem is stated on page 24 of LMCS. Boole's proofs, as explained in the historical remarks, were pretty `wild'. Here is a modern version.
Theorem [Elimination Theorem]
The most general equation that follows from
is obtained by setting
or, more briefly,
where ranges over sequences of 1's and 0's.
Proof.
First we show , for the stated choice of F.
By expanding E about
(see Problem 4.4 on page 25)
we have
where . Then from
follows
and thus
Taking the union of these two gives
and thus
Now repeat the above steps, except this time expand about ,
using the last equation which has the
form
, to obtain
where . Continuing one arrives at the desired
conclusion that
, as defined above, is 0.
For the converse suppose follows from
. Let
be any
-constituent of H. Then for any
-constituent
we have
is an
-constituent
of H. But then, by Theorem 4.5, LK must be an
-constituent
of E as
follows from
.
Then
has K as a
-constituent,
i.e.,
.
So K is a
-constituent of F as F is
defined to be the intersection over the L's of
the
's.
So every -constituent of H is also
a
-constituent of F. This means,
by Theorem 4.5, that
follows from
, so
is a more general conclusion than
.