FINITE BOOLEAN ALGEBRA
1. Deconstructing Boolean algebras with atoms.
Let B = <B,≤,¬,∧,∨,0,1> be a Boolean algebra and c ∈ B.
The ideal generated by c, (c], is:
(c] = {b ∈ B: b ≤ c}
The filter generated by c, [c), is:
[c) = {b ∈ B: c ≤ b}
The ideal-relativization of B to c, (c], is the structure:
(c] = <(c],≤(c],¬(c],∧(c],∨(c],0(c],1(c]>, where:
1. ≤(c] = ≤⌠(c]
2. ¬(c] = {<x,¬x ∧ c>: x ∈ (c]}
3. ∧(c] = ∧⌠(c]
4. ∨(c] = ∨⌠(c]
5. 0(c] = 0
6. 1(c] = c
THEOREM 1: (c] is a Boolean algebra.
PROOF:
1. <(c],≤(c],∧(c],∨(c],0(c]> is a substructure of <B,≤,∧,∨,0>: namely, if a,b ∈ (c], then
a,b ≤ c, hence (a ∧ b) ≤ c and (a ∨ b) ≤ c, hence (a ∧ b) ∈ (c] and (a ∨ b) ∈ (c].
Further 0(c] = 0.
2. <(c],≤(c],∧(c],∨(c],0(c],1(c]> is bounded by 0(c] and 1(c]. We have seen already that 0(c]
is the minimum of (c]. 1(c] = c, and c is, obviously, the maximum of (c].
3. We have proved under 1. that <(c],≤(c],∧(c],∨(c]> is a sublattice of <B,≤,∧,∨>. Since
<B,≤,∧,∨> is distributive, it follows that <(c],≤(c],∧(c],∨(c]> is distributive (since the
class of distributive lattices is closed under substructure).
4. Thus, we have proved that (c] is a bounded distributive lattice. So we only need
to prove that (c] is complemented, i.e. that ¬(c] is complementation on (c].
First: (c] is closed under ¬(c]. This is obvious, since obviously ¬x ∧ c ≤ c, for any
x ∈ B, hence also for any x ≤ c.
Secondly, ¬(c] respects the laws of 0(c] and 1(c]:
Let a ∈ (c]. a ∧(c] ¬(c](a) = a ∧ (¬a ∧ c) = (a ∧ ¬a) ∧ (a ∧ c) = 0 ∧ (a ∧ c) = 0
a ∨(c] ¬(c](a) = a ∨ (¬a ∧ c) = (a ∨ ¬a) ∧ (a ∨ c) = 1 ∧ (a ∨ c) = a ∨ c = c
Thus, indeed ¬(c] is complementation on (c]. ◄
Note: except for the trivial case where c is 1, (c] is not a sub-Boolean algebra of B,
because 1 is not preserved. It is a Boolean algebra on a subset of B.
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The filter-relativization of B to c, [c), is the structure:
[c) = <B[c),≤ [c),¬ [c),∧ [c),∨ [c),0[c),1[c)>, where:
1. ≤[c) = ≤⌠B[c)
2. ¬[c) = {<x,¬x ∨ c>: x ∈ [c)}
3. ∧[c) = ∧⌠[c)
4. ∨[c) = ∨⌠[c)
5. 0[c) = c
6. 1[c) = 1
THEOREM 2: [c) is a Boolean algebra.
PROOF:
1. <[c),≤ [c),∧ [c),∨ [c),1[c)> is a substructure of <B,≤,∧,∨,1>: namely, if a,b ∈ [c), then
c ≤ a,b, hence c ≤ (a ∧ b) and c ≤ (a ∨ b), hence (a ∧ b) ∈ [c) and (a ∨ b) ∈ [c).
Further 1[c) = 1.
2. <[c),≤ [c),∧ [c),∨ [c),0[c),1[c)> is bounded by 0[c) and 1[c). We have seen already that
1[c)is the maximum of [c). 0[c) = c, and c is, obviously, the mimimum of [c).
3. We have proved under 1. that <[c),≤ [c),∧ [c),∨ [c)> is a sublattice of <B,≤,∧,∨>.
Since <B,≤,∧,∨> is distributive, it follows that <[c),≤ [c),∧ [c),∨ [c)> is distributive
(since the class of distributive lattices is closed under substructure).
4. Thus, we have proved that [c) is a bounded distributive lattice. So we only need
to prove that [c) is complemented, i.e. that ¬[c) is complementation on [c).
First: [c) is closed under ¬[c). This is obvious, since obviously c ≤ ¬x ∨ c, for any
x ∈ B, hence also for any x such that c ≤ x.
Secondly, ¬[c) respects the laws of 0[c) and 1[c):
Let a ∈ [c). a ∧[c) ¬[c) (a) = a ∧ (¬a ∨ c) = (a ∧ ¬a) ∨ (a ∧ c) = 0 ∨ (a ∧ c) = a ∧ c
=c
a ∨[c) ¬[c) (a) = a ∨ (¬a ∨ c) = (a ∨ ¬a) ∨ (a ∨ c) = 1 ∨ (a ∨ c) = 1
Thus, indeed ¬[c) is complementation on [c). ◄
LEMMA 3: If c ≠ 1 then (c] ∩ [¬c) = Ø
PROOF: Let x ∈ (c] ∩ [¬c). Then x ≤ c and ¬c ≤ x. Then x ≤ c and and ¬x ≤ c.
Then x ∨ ¬x ≤ c, hence c = 1. ◄
THEOREM 4: (c] and [¬c) are isomorphic.
PROOF:
If c =1, then (c] = [¬c) = B. So clearly they are isomorphic.
So let c ≠ 1.
We define: h: (c] → [¬c) by:
for every x ∈ (c]: h(x) = x ∨ ¬c.
1. Since for every x ∈ B, ¬c ≤ x ∨ c, also for every x ∈ (c]: ¬c ≤ x ∨ c. Hence for
every x ∈ (c]: h(x) ∈ [¬c), hence h is indeed a function from (c] into [¬c).
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2. Let y ∈ [¬c). Then ¬c ≤ y. Then ¬y ≤ c, hence ¬y ∈ (c]. Take the relative
complement of ¬y in (c]: ¬¬y ∧ c. This is y ∧ c ∈ (c].
h(y ∧ c) = (y ∧ c) ∨ ¬c = (y ∨ ¬c) ∧ (c ∨ ¬c) = (y ∨ ¬c) ∧ 1 = y ∨ ¬c = y.
Hence h is onto.
3. Let h(x1) = h(x2). Then x1 ∨ ¬c = x2 ∨ ¬c. Then ¬(x1 ∨ ¬c) = ¬( x2 ∨ ¬c).
Then ¬x1 ∧ c = ¬x2 ∧ c. Since these are the relative complements of x1 and x2 in
Boolean algebra (c], it follows that x1 = x2.
Hence h is one-one.
4. h(0) = 0 ∨ ¬c = ¬c.
h(c) = c ∨ ¬c = 1
h(a ∧ b) = (a ∧ b) ∨ ¬c = (a ∨ ¬c) ∧ (b ∨ ¬c) = h(a) ∧ h(b)
h(a ∨ b) = (a ∨ b) ∨ ¬c = (a ∨ ¬c) ∨ (b ∨ ¬c) = h(a) ∨ h(b)
h(¬(c](a)) = h(¬a ∧ c) = (¬a ∧ c) ∨ ¬c = ¬(a ∨ ¬c) ∨ ¬c = ¬[¬c)(a ∨ ¬c) =
¬[¬c)(h(a)).
Thus, indeed, h is an isomorphism. ◄
LEMMA 5: If a is an atom in B, then for every x ∈ B-{0}: a ≤ x or a ≤ ¬x.
PROOF: Let a be an atom in B. Suppose that ¬(a ≤ x). Then (a ∧ x) ≠ a. But
a ∧ x ≤ a. Since a is is an atom, that means that a ∧ x = 0. But that means that a ≤ ¬x.
◄
CORROLLARY 6: If a is an atom in B, then (¬a] ∪ [a) = B.
PROOF: This follows from lemma 5: Let x ∈ B and x ∉ [a). Then ¬(a ≤ x). Hence
by lemma 5 a ≤ ¬x, and that means that x ≤ ¬a, hence x ∈ (¬a]. ◄
All this has the following consequence for finite Boolean algebras:
THEOREM 7: Let B be a finite Boolean algebra. Then |B| = 2n, for some n≥0.
PROOF:
If |B| = 1 then |B| = 20. If |B| = 2 then |B| = 21.
Let |B| > 2. We define for B a decomposition tree DEC(B) in the following way:
top(DEC(B)) = <B,a0,0>, with ¬a0 an atom in B.
for every node <A,a,n> in DEC(B): if |A|>2 then
daughters(<A,a,n>) = {<(a],a1,n+1>,<[¬a),a2,n+1>}, with ¬a1 an atom in (a]
and ¬a2 an atom in [¬a).
Let <A1,aA1,k+1>, <A,aA,k> ∈ DEC(B) and let <A1,aA1,k+1> be a daughter of
<A,aA,k>. Then |A| = 2 × |A1|.
Obviously, this means that for any node <A,aA,k> ∈ DEC(B): |B| = 2k × |A|
This means that if <A1,aA1,k>, <A2,aA2,k> ∈ DEC(B), then |A1|=|A2|.
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And that means that if <A1,aA1,k>, <A2,aA2,k> ∈ DEC(B), either both A1 and A2
decompose (if |A1|>2), or neither do (If |a1|≤2).
Thus for any k such that some <A,aA,k> ∈ DEC(B): all nodes <A,aA,k> ∈ DEC(B)
decompose, or none do.
This means that for some k > 0: leave(DEC(B)) = {<A,aA,k>: <A,aA,k> ∈ DEC(B)}
(all leaves have the same level, and hence the same cardinality.)
Let <A,aA,k> ∈ leave(DEC(B). Then it follows that |B| = 2k × |A|.
Since <A,aA,k> ∈ leave(DEC(B), |A|≤2.
For some <C,aC,k-1> ∈ DEC(B), <A,aA,k> is the daughter of <C,aC,k-1>, hence
|C| = 2 × |A|, |C| > 2 and |A| ≤ 2. This means that |A|=2.
Hence |B| = 2k+1. Hence for some n>1: |B| = 2n.
We have now proved that for every finite Boolean algebra B |B| = 2n for some n≥0.
◄
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2. Constructing product Boolean algebras.
Let A and B be Boolean algebras.
The product of A and B, A × B, is given by:
A × B = <B×,≤×, ¬×, ∧×, ∨×, 0×, 1×> where
1. B× = A × B
2. ≤x = {<<a1,b1>:,<a2,b2>>: a1 ≤A a2 and b1 ≤B b2}
3. For every <a,b> ∈ A × B:¬×(<a,b>) = <¬Aa,¬Bb>
4. For every <a1,b1>,<a2,b2> ∈ A × B:
<a1,b1> ∧× <a2,b2> = <a1 ∧A a2,b1 ∧B b2>
5. For every <a1,b1>,<a2,b2> ∈ A × B:
<a1,b1> ∨× <a2,b2> = <a1 ∨A a2,b1 ∨B b2>
6. 0× = <0A,0B>
7. 1x = <1A,1B>
THEOREM 8: A × B is a Boolean algebra.
PROOF:
1. ≤× is a partial order.
reflexive:
Since for every a ∈ A: a ≤A a and for every b ∈ B: b ≤B b,
for every <a,b> ∈ A × B: <a,b> ≤x <a,b>.
antisymmetric:
Let <a1,b1> ≤× <a2,b2> and <a2,b2> ≤× <a1,b2>.
Then a1 ≤A a2 and b1 ≤B b2 and a2 ≤A a1 and b2 ≤B b1,
hence a1 = a2 and b1 = b2, hence <a1,b2> = <a2,b2>
transitive:
Let <a1,b1> ≤× <a2,b2> and <a2,b2> ≤× <a3,b3>.
Then a1 ≤A a2 and b1 ≤B b2 and a2 ≤A a3 and b2 ≤B b3,
hence a1 ≤A a3 and b1 ≤B b3, hence <a1,b1> ≤× <a3,b3>
2. <a1,b1> ∧× <a2,b2> = <a1 ∧A a2,b1 ∧B b2>
a1 ∧A a2 ≤A a1, a1 ∧A a2 ≤A a2, b1 ∧B b2 ≤B b1, b1 ∧B b2 ≤B b2, hence
<a1,b1> ∧× <a2,b2> ≤× <a1,b1> and <a1,b1> ∧× <a2,b2> ≤× <a2,b2>.
Let <a,b> ≤× <a1,b1> and <a,b> ≤× <a2,b2>.
Then a ≤A a1 and b ≤B b1 and a ≤A a2 and b ≤B b2, hence
a ≤A a1 ∧A a2 and b ≤B b1 ∧B b2, hence
<a,b> ≤× <a1,b1> ∧× <a2,b2>.
Hence ∧× is meet in ≤×.
3. We show that ∨× is join in ≤× by a similar argument.
4. 0× = <0A,0B>. Since for every a ∈ A 0A ≤A a and for every b ∈ B 0B ≤B b,
for every <a,b> ∈ A × B <0A,0B> ≤× <a,b>. Hence 0× is the minumum under ≤×.
Similarly 1× is the maximum under ≤A
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So A × B is a bounded lattice.
5. <a1,b1> ∧× (<a2,b2> ∨× <a3,b3>) =
<a1 ∧A (a2 ∨A a3),b1 ∧B (b2 ∨B b3)> =
<(a1 ∧A a2) ∨A (a1 ∧A a3),(b1 ∧B b2) ∨B (b1 ∧B b3)> =
(<a1 ∧A a2,b1 ∧B b2> ∨× <a1 ∧A a3,b1 ∧B b3> =
(<a1,b1> ∧× <a2,b2>) ∨× (<a1,b1> ∧x <a2,b2>)
So A × B is distributive.
6. ¬× satisfies the laws of 0× and 1×:
<a,b> ∧× ¬×(<a,b>) = <a,b> ∧× <¬Aa,¬Bb> = <a ∧A ¬Aa,b ∧B ¬Bb> =
<0A,0B> = 0×.
<a,b> ∨× ¬×(<a,b>) = <a,b> ∨× <¬Aa,¬Bb> = <a ∨A ¬Aa,b ∨B ¬Bb> =
<1A,1B> = 1×.
So A × B is a Boolean algebra. ◄
Let B1 and B2 be isomorphic Boolean algebras such that B1 ∩ B2 = Ø, and let h be an
isomorphism between B1 and B2.
We definite the product of B1 and B2 under h, Bh1+2:
.
Bh1+2 = < Bh1+2,≤1+2,¬1+2,∧1+2,∨1+2,01+2,11+2> where:
1. B1+2 = B1 ∪ B2.
2. ≤1+2 = ≤1 ∪ ≤2 ∪ {<b1,b2>: h(b1) ≤2 b2}
3. ¬1+2 is defined by:
if b ∈ B1
¬2(h(b))
For all b ∈ B1+2: ¬1+2(b) =
if b ∈ B2
¬1(h-1(b))
4. ∧1+2 is defined by:
a ∧1 b
if a,b ∈ B1
if a,b ∈ B2
For all a,b ∈ B1+2: a ∧1+2 b = a ∧2 b
a ∧1 h-1(b) if a ∈ B1 and b ∈ B2
5. ∨1+2 is defined by:
a ∨1 b
if a,b ∈ B1
if a,b ∈ B2
For all a,b ∈ B1+2: a ∨1+2 b = a ∨2 b
h(a) ∨1 b if a ∈ B1 and b ∈ B2
6. 01+2 = 01.
7. 11+2 = 11.
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THEOREM 9: Bh1+2 is a Boolean algebra.
PROOF:
1. ≤1+2 is a partial order.
≤1+2 is reflexive:
If a ∈ B1: a ≤1 a, hence, a ≤1+2 a
If a ∈ B2: a ≤2 a, hence, a ≤1+2 a
≤1+2 is antisymmetric.
Let a ≤1+2 b and b ≤1+2 a. This is only possible if a,b ∈ B1 or a,b ∈ B2.
In the first case a ≤1 b and b ≤1 a, hence a=b.
In the second case a ≤2 b and b ≤1 a, hence a=b.
≤1+2 is transitive.
Let a ≤1+2 b and b ≤1+2 c
If a,b,c ∈ B1, then a ≤1 b and b ≤1 c, hence a ≤1 c, and a ≤1+2 c
If a,b,c ∈ B2, then a ≤2 b and b ≤2 c, hence a ≤2 c, and a ≤1+2 c
If a ∈ B1 and b,c ∈ B2, then h(a) ≤2 b and b ≤2 c.
Then h(a) ≤2 c and a ≤1+2 c.
If a,b ∈ B1 and c ∈ B2, then a ≤1 b and h(b) ≤2 c.
Since h is an isomorphism, this means that h(a) ≤2 h(b), and hence
h(a) ≤2 c. Hence a ≤1+2 c.
2. ∧1+2 is meet under ≤1+2.
If a,b ∈ B1: a ∧1+2 b = a ∧1 b, which is meet under ≤1, and ≤1 = ≤1+2⌠B1.
If a,b ∈ B2: a ∧1+2 b = a ∧2 b, which is meet under ≤2, and ≤2 = ≤1+2⌠B2.
If a ∈ B1 and b ∈ B2, then a ∧1+2 b = a ∧1 h-1(b).
a ∧1 h-1(b) ≤1 a and a ∧1 h-1(b) ≤1 h-1(b).
By definition of ≤1+2, h-1(b) ≤1+2 h(h-1(b)).
So h-1(b) ≤1+2 b. Then a ∧1 h-1(b) ≤1+2 b.
This means that a ∧1+2 b ≤1+2 a and a ∧1+2 b ≤1+2 b.
If x ≤1+2 a and x ≤1+2 b, then x ≤1 a and h(x) ≤2 b.
Since h is an isomorphism, then h-1(h(x)) ≤1 h-1(b), i.e. x ≤1 h-1(b).
then x ≤1 a ∧1 h-1(b) Hence x ≤1+2 a ∧1+2 b.
So indeed ∧1+2 is meet under ≤1+2.
3.. ∨1+2 is join under ≤1+2.
If a,b ∈ B1: a ∨1+2 b = a ∨1 b, which is join under ≤1, and ≤1 = ≤1+2⌠B1.
If a,b ∈ B2: a ∨1+2 b = a ∨2 b, which is join under ≤2, and ≤2 = ≤1+2⌠B2.
If a ∈ B1 and b ∈ B2, then a ∨1+2 b = h(a) ∨2 b.
b ≤2 h(a) ∨2 b and h(a) ≤2 h(a) ∨2 b.
As we have seen a ≤1+2 h(a), hence a ≤1+2 h(a) ∨2 b.
So a ≤1+2 a ∨1+2 b and b ≤1+2 a ∨1+2 b.
If a ≤1+2 x and b ≤1+2 x, then h(a) ≤1+2 x, hence h(a) ∨2 b ≤2 x.
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Hence a ∨1+2 b ≤1+2 x.
So indeed ∨1+2 is join under ≤1+2.
4. 01+2 = 01.
If a ∈ B1, 01 ≤1 a. hence 01+2 ≤1+2 a.
h is an isomorphism, so h(01) = 02.
If a ∈ B2, then h(01 ≤2 a, hence 01+2 ≤1+2 a.
So indeed 01+2 is the minimum under ≤1+2.
Similarly, 11+2 is the maximum under ≤1+2.
We have proved so far that B1+2h is a bounded lattice.
5. Distributivity: a ∧1+2 (b ∨1+2 c) = (a ∧1+2 b) ∨1+2 (a ∧1+2 c)
a. Let a,b,c ∈ B1 or a,b,c ∈B2, then distributivity follows from distributivity of ∧1 and
∨1 and of ∧2 and ∨2.
b. Let a ∈ B1 and b,c ∈ B2
a ∧1+2 (b ∨1+2 c) = a ∧1 h-1(b ∨2 c) = a ∧1 (h-1(b) ∨1 h-1(c)) =
(a ∧1 h-1(b)) ∨1 (a ∧1 h-1(c)) = (a ∧1+2 b) ∨1+2 (a ∧1+2 c)
c. Let b ∈ B1 and a,c ∈ B2
a ∧1+2 (b ∨1+2 c) = a ∧2 (h(b) ∨2 c) = (a ∧2 h(b)) ∨2 (a ∧2 c)
If a ∈ B2 and b ∈ B1, then a ∧1+2 b = h-1(a) ∧1 b.
Then h(a ∧1+2 b) = h(h-1(a) ∧1 b) = h(h-1(a)) ∧2 h(b) = a ∧2 h(b)
Hence (a ∧2 h(b)) ∨2 (a ∧2 c) = h(a ∧1+2 b) ∨2 (a ∧2 c) = (a ∧1+2 b) ∨1+2 (a ∧1+2 c)
d. Let b,c ∈ B1 and a ∈ B2
a ∧1+2 (b ∨1+2 c) = a ∧1+2 (b ∨1 c) = h-1(a) ∧1 (b ∨1 c) =
(b ∧1 h-1(a)) ∨1 (c ∧1 h-1(a)) = (a ∧1+2 b) ∨1+2 (a ∧1+2 c)
e. Let a,b ∈ B1 and c ∈ B2
a ∧1+2 (b ∨1+2 c) = a ∧1 ( h-1(b ∨1+2 c))
If b ∈ B1 and c ∈ B2 , then b ∨1+2 c = h(b) ∨2 c
Hence h-1(b ∨1+2 c) = h-1(h(b) ∨2 c) = h-1(h(b)) ∨1 h-1(c) = b ∨1 h-1(c)
So a ∧1 ( h-1(b ∨1+2 c)) = a ∧1 (b ∨1 h-1(c)) = (a ∧1 b) ∨1 (a ∧1 h-1(c)) =
(a ∧1+2 b) ∨1+2 (a ∧1+2 c).
These are all the relevant cases, so B1+2h is a distributive bounded lattice.
5.. ¬1+2 satisfies the laws of 01+2 and 11+2.
If a ∈ B1,
a ∧1+2 ¬1+2(a) = a ∧1 h-1(¬1+2(a)) = a ∧1 h-1(¬2(h(a))) = a ∧1 h-1(h(¬1a)) = a ∧1 ¬1a =
01 = 01+2
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a ∨1+2 ¬1+2(a) = h(a) ∨2 ¬1+2(a) = h(a) ∨2 ¬2(h(a) = 12 = 11+2.
If a ∈ B2,
a ∧1+2 ¬1+2(a) = h-1(a) ∧1 ¬1+2(a) = h-1(a) ∧1 ¬1(h-1(a) = 01 = 01+2
a ∨1+2 ¬1+2(a) = a ∨2 h(¬1+2(a)) = a ∨2 h(¬1(h-1(a)) = a ∨2 ¬2(h(h-1(a)) = a ∨2 ¬2a =
12 = 11+2.
Thus B1+2h is a Boolean algebra. ◄
THEOREM 10:Let B1 and B2 be isomorphic Boolean algebras such that B1 ∩ B2= Ø,
and let h be an isomorphism between B1 and B2..
Let {0,1} be a Boolean algebra of cardinality 2.
B1+2h is isomorphic to B1 × {0,1}
PROOF:
We define function k from B1 ∪ B2 into B1 × {0,1):
For all x ∈ B1: k(x) = <x,0>
For all x ∈ B2: k(x) = <h-1(x),1>
1. Since h is an isomorphism between B1 and B2, and B1 ∩ B2 = Ø, k is obviously a
bijection between B1 ∪ B2 and B1 × {0,1).
2. If x ∈ B1, k(¬1+2(x)) = <h-1(¬1+2(x)),1> = <h-1(¬2(h(x)), ¬{0,1}0> =
<h-1(h(¬1(x))), ¬{0,1}0> = <¬1(x), ¬{0,1}0> = ¬×<x,0>
3. k(01+2) = k(01) = <01,0> = 0×.
k(11+2) = k(12) = <h-1(12),1> = <11,1> = 1×.
4. k preserves meet:
If a,b ∈ B1 then k(a ∧1+2 b) = k(a ∧1 b) = <a ∧1 b,0> = <a,0> ∧× <b,0> =
k(a) ∧× k(b).
If a,b ∈ B2 then k(a ∧1+2 b) = k(a ∧2 b) = <h-1(a ∧2 b),1> = <h-1(a) ∧1 h-1(b),1> =
<h-1(a),1> ∧× <h-1(b),1> = k(a) ∧× k(b).
If a ∈ B1 and b ∈ B2 then k(a ∧1+2 b) = k(a ∧1 h-1(b)) = <a ∧1 h-1(b),0> =
<a ∧1 h-1(b),0 ∧{0,1} 1>
<a,0> ∧× <h-1(b),1> = k(a) ∧× k(b).
5. k preserves join:
If a,b ∈ B1 then k(a ∨1+2 b) = k(a ∨1 b) = <a ∨1 b,0> = <a,0> ∨× <b,0> =
k(a) ∨× k(b).
If a,b ∈ B2 then k(a ∨1+2 b) = k(a ∨2 b) = <h-1(a ∨2 b),1> = <h-1(a) ∨1 h-1(b),1> =
<h-1(a),1> ∨× <h-1(b),1> = k(a) ∨× k(b).
9
If a ∈ B1 and b ∈ B2 then k(a ∨1+2 b) = k(h(a) ∨2 b) = <h-1(h(a) ∨2 b),1> =
<a ∨1 h-1(b),1> = <a ∨1 h-1(b),0 ∨{0,1} 1> = <a,0> ∨× <h-1(b),1> = k(a) ∨× k(b).
Thus indeed k is an isomorphism. ◄
THEOREM 11: Let B be a Boolean algebra of cardinality larger than 2 and let ¬a be
an atom in B.
Let h: (a] → [¬a) be the isomorphism defined by:
for every x ∈ (a]: h(x) = x ∨ ¬a.
Then B(a]+[¬a)h = B.
PROOF:
1. B(a]+[¬a) = (a]+[¬a) = B.
2. ≤(a]+[¬a) = ≤B.
a. Let <x,y> ∈ ≤(a]+[¬a).
Either x,y ∈ (a], then x ≤B y, or x,y ∈ [¬a), then x ≤B y, or x ∈ (a] and y ∈ [¬a) and
h(x) ≤[¬a) y. Since h(x) = x ∨B ¬a, then obviously x ≤B y.
So in all cases <x,y>∈ ≤B.
b. Let <x,y> ∈ ≤B, i.e. x ≤B y.
Either x,y ∈ (a], then <x,y> ∈ ≤(a]+[¬a), or x,y ∈ [¬a), then <x,y> ∈ ≤(a]+[¬a).
It can't be the case that y ∈ (a] and x ∈ [¬a), because then y ∈ [¬a), but then the
intersection of (a] and [¬a) would not be empty, and it is.
This leaves only the case that x ∈ (a] and y ∈ [¬a).
Now x ≤B y and ¬a ≤B y, hence x ∨B ¬a ≤B y. But h(x) = x ∨B ¬a. Hence
<x,y> ∈ ≤(a]+[¬a).
Thus, indeed ≤(a]+[¬a) = ≤B.
This means that B(a]+[¬a)h and B are the same partial order. That means, of course,
that they have identical joins and meets, and this means that they are the same
bounded distributive lattice. Since in a bounded distributive lattice, each element has
a unique complement and since ¬(a]+[¬a) and ¬B map every element onto its
complement, ¬(a]+[¬a) = ¬B. Hence, indeed, B(a]+[¬a)h and B are the same Boolean
algebra. ◄
10
THEOREM 12: Let B1 and B2 be Boolean algebras, ¬a1 an atom in B1 and ¬a2 an
atom in B2, and let (a1] be isomorphic to (a2]. Then B1 and B2 are isomorphic.
PROOF:
Let h1 be the isomorphism between (a1] and [¬a1) defined by:
for all x ∈ (a1]: h1(x) = x ∨B1 ¬B1(a1)
Let h2 be the isomorphism between (a2] and [¬a2) defined by:
for all x ∈ (a2]: h2(x) = x ∨B2 ¬B2(a2)
Let k the isomorphism between (a1] and (a2].
Let {0,1} be a two element Boolean algebra.
B1 = B(a1] + [¬a1)h1 and B2 = B(a2] + [¬a2)h2, by theorem11.
B1 is isomorphic to (a1] × {0,1} and B2 is isomorphic to (a2] × {0,1}, by theorem 10.
Define g: (a1] × {0,1} → (a2] × {0,1} by:
for every <a,b> ∈ (a1] × {0,1}: g(<a,b>) = <k(a),b>.
It is straightforward to prove that g is an isomorphism between B1 and B2. ◄
All this has the following consequences for finite Boolean algebras:
THEOREM 13: Any two finite Boolean algebras of the same cardinality are
isomorphic.
PROOF:
1. Obviously, up to isomorphism, there is only one Boolean algebra of cardinality 1 or
cardinality 2. Up to isomorphism, there is only one partial order of cardinality 1,
hence also only one Boolean algebra. Up to isomorphism there are two partial orders
of cardinality 2: <{0,1},{<0,0>,<1,1>}> and <{0,1},{<0,0>,<0,1>,<1,1>}>. Only the
second is a lattice and a Boolean algebra.
2. If all Boolean algebras of cardinality 2n, n>0 are isomorphic, then all Boolean
algebras of cardinality 2n+1 are isomorphic.
This follows from theorem 12. Let B1 and B2 be Boolean algebras of cardinality 2n+1,
and assume that all Boolean algebras of cardinality 2n are isomorphic.
Let ¬a1 be an atom in B1 and ¬a2 be an atom in B2. (a1] and (a2] are Boolean algebras
of cardinality 2n, hence, by assumption they are isomorphic. Then, by theorem12, B1
and B2 are isomorphic.
1 and 2 together prove that any two finite Boolean algebras of the same cardinality are
isomorphic. ◄
CORROLLARY 14: Up to isomorphism the finite Boolean algebras are exactly the
finite powerset Boolean algebras.
PROOF:
For every n ≥ 1, if |X|=n then <pow(X),-,∩,∪,X,Ø> is a powerset Boolean algebra of
cardinality 2n. By theorem13, every Boolean algebra of cardinality 2n is isomorphic
to it. ◄
11
So, we can construct every finite Boolean algebra as a powerset Boolean algebra.
We can also use the product construction under an isomorphism to construct all finite
Boolean algebras.
Cardinality 1:
o1
a point in 0-dimensional space.
Cardinality 2:
Take two non-overlapping Boolean algebras of cardinality 1 and an isomorphism, and
construct the product:
o1
+
o2
o1
+
{<1,2>}
→
a point moved along a new dimension:
a line in 1-dimensional space.
o2
Cardinality 4:
Take two non-overlapping Boolean algebras of cardinality 2 and an isomorphism, and
construct the product:
o2
+
o1
o4
+
→
{<1,3>,<2,4>}
o3
o4
o2
A line moved along a new dimension.
A square in 2-dimensional space.
o3
o1
Cardinality 8:
Take two non-overlapping Boolean algebras of cardinality 4 and an isomorphism, and
construct the product:
o4
o2
o8 + {<1,5>,<2,6>,<3,7>,<4,8>} →
+
o3
o6
o7
o1
o5
12
o8
A square moved along a new dimension.
A cube in 3-dimensional space.
6o
o4
o7
2o
o5
o3
o1
Cardinality 16:
Take two non-overlapping Boolean algebras of cardinality 8 and an isomorphism, and
construct the product:
8
o
+
16
o
6
o
o4
7
o
14
o
o 12
15
o
o
2
o
5
o
3
o
10
o
13
o
11
o
1
+
o
9
{<1,9>,<2,10>,<3,11>,<4,12>,<5,13>,<6,14>,<7,15>,<8,16>}
→
16
o
8
o
14
o
6
o
o4
7
o
o
2
o
5
o
3
o 12
15
o
o
o
10
13
o
9
o
1
A cube moved along a new dimension.
13
o
11
A 4-dimensional object:
16
o
8
o
14
o
6
o
o4
7
o
o
2
o
5
o
3
o 12
15
o
o
o
10
13
o
9
o
1
14
o
11