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4.3 More Results about Boolean Algebras

63

algebra’s unit element on every algebraic interpretation such that all of the members of evaluate to unit—coincide with entailments under the standard semantics.

Interestingly, we also have the following result:

Result 4.3: The formulas of propositional logic that are tautologies for every Boolean algebraic semantics are exactly the tautologies under the standard semantics based on T and F.

Proof: Because the standard semantics based on T and F is a Boolean algebraic semantics, every formula that is a tautology for every Boolean algebraic semantics must therefore be a tautology under the standard semantics.

For the converse, we can draw on the fact that the system CLA presented in Chapter 2 is a complete axiomatization for propositional logic under the standard semantics; that is, every tautology under the standard semantics is a theorem of CLA. Then we shall only need to prove that every theorem of CLA is a tautology under every Boolean algebraic semantics—and this we can do by establishing that the three axiom schemata of CLA are tautologous under every Boolean algebraic semantics and that Modus Ponens preserves this property. Specifically, we need to establish that x (y x) = unit in every Boolean algebra (the left-hand side is the algebraic formula corresponding to CLA1), and that (x (y z)) ((x y) (x z)) = unit (CLA2) and (x y ) (y x) = unit (CLA3)—again, the reader will be asked to prove these equations in the exercises for Section 4.3, as well as the claim that Modus Ponens preserves tautologousness in any Boolean algebra.

Apropos of Result 4.3, our only examples of Boolean algebras so far have been two-valued. As a final and more general example, we note that given any nonempty set S there is a standard way to generate a Boolean algebra using set-theoretic operations. Let and denote set-theoretic union and intersection, let denote complementation relative to the set S (i.e., for any subset X of S, X will be the set of elements of S that are not members of X), and let P(S) denote the power set of S, that is, the set of all subsets of S. The structure <P(S), , , , S, Ø> is then a Boolean algebra. To verify this we need to check that each of conditions i–vii holds for this structure. For any sets X and Y, X Y = Y X and X Y = Y X, so condition i is met. Moving to condition vi, we note that for any set X S, X Ø = X and X S = X. The interested reader can verify that all of the other conditions hold as well. Thus it is not surprising that the two-valued semantics based on {5} and Ø that we presented for classical logic is a Boolean algebra.

4.3 More Results about Boolean Algebras

Not all of the seven conditions of Section 4.2 are needed to define Boolean algebras. Conditions ii, iii, and iv are derivable from the others (we explain later why we have

64

Alternative Semantics for Truth-Values and Truth-Functions

nevertheless included them in the definition). Here is how the first idempotence condition (iii) can be derived:

x x = (x x) unit

(identity for meet)

= (x x) (x x )

(complementation)

= x (x x )

(distribution)

= x zero

(complementation)

= x

(identity for join)

To derive the first absorption condition we first derive the law unit x = unit, which we call unit consumption:

unit x = (unit x) unit

(identity for meet)

= unit (x unit)

(commutation, twice)

= (x x ) (x unit)

(complementation)

= x (x unit)

(distribution)

= x x

(identity for meet)

= unit

(complementation)

and then we use unit consumption to derive the first absorption condition (iv):

x (x y) = (x unit) (x y)

(identity for meet)

= x (unit y)

(distribution)

= x unit

(unit consumption)

= x

(identity for meet)

To derive the first association condition we begin with the following, in which the right-hand side ((x y) z) of the formula is associated to (x (y z)):3

x ((x y) z) = (x (x y)) (x z)

(distribution)

= x (x z)

(absorption)

= x

(absorption)

= x (x (y z))

(absorption)

= (x x) (x (y z))

(idempotence)

= x (x (y z))

(distribution)

(Note that we have already shown how to derive idempotence from conditions i and v–vii on Boolean algebras.) Next we show that the same result can be derived when we replace x on the left-hand side with its complement:

3This proof, which admittedly is tricky and not at all the obvious way to go, is based on an outline in Stoll (1961, p. 253).

4.3 More Results about Boolean Algebras

65

x ((x y) z) = (x (x y)) (x z)

(distribution)

= ((x x) (x y)) (x z)

(distribution)

= (zero (x y)) (x z)

(complementation)

= (x y) (x z)

(identity for join)

= x (y z)

(distribution)

= zero (x (y z))

(identity for join)

= (x x) (x (y z))

(complementation)

= x (x (y z))

(distribution)

Because of these identities we can assert

 

(x ((x y) z)) (x ((x y) z)) = (x (x (y z))) (x (x (y z)))

and from this we can derive

(((x y) z) x) (((x y) z) x ) = ((x (y z)) x) ((x (y z)) x )

by commutation on both sides, then

((x y) z) (x x ) = (x (y z)) (x x )

by distribution, then

((x y) z) unit = (x (y z)) unit

by complementation, and finally

(x y) z = x (y z)

by identity for meet. This last formula is the first associative condition for Boolean algebras.

The second idempotence, absorption, and association conditions can be similarly derived from conditions i and v–vii and are left as an exercise.

We have included the derivable conditions ii–iv in the definition of Boolean algebras to make clear the connection between Boolean algebras and another type of algebraic structure called a lattice. A lattice is a structure <L, , > for which conditions i–iv of Boolean algebras hold. If condition v also holds, the lattice is said to be distributed. If the set L contains two elements that can fill the role of zero and unit in the identity condition vi, then those elements will be the zero and unit elements and the lattice is accordingly said to be bounded or, alternatively, to contain zero and unit elements. If in addition corresponding to each member x of L there is a member y of L such that x y = unit and x y = zero, then each such x and y are called complements (because y functions as x in the complementation condition for Boolean algebras) and the lattice is said to be complemented. So a

66 Alternative Semantics for Truth-Values and Truth-Functions

Boolean algebra is a special type of lattice, namely, a complemented distributive lattice with zero and unit elements.

Each of the conditions defining Boolean algebras is specified by a pair of equations, where the second equation results from the first by exchanging the meet and join operations in each formula and exchanging zero and unit. Such pairs are called duals, and the fact that the conditions come as dual pairs entails that in a Boolean algebra the dual of any provable equation is also provable (by using the same proof, except that each formula is replaced by its dual). So for example, since we can show

that unit

=

zero as follows:

 

 

 

unit

=

zero

 

zero

(complementation)

 

 

 

= zero

zero

(commutation)

 

=

zero

 

 

 

(identity for join)

 

 

 

 

 

we know that zero

=

unit is also provable using the dual formulas:

 

zero

=

unit

unit

(complementation)

 

 

 

= unit

unit

(commutation)

 

=

unit

 

 

 

(identity for meet)

 

 

 

 

 

The following laws also hold in every Boolean algebra:

Double Negation Law: x = x

DeMorgan’s Laws: (x y) = x y , and (x y) = x y

We first derive Double Negation:

 

x = x zero

(identity for join)

= x (x x )

(complementation)

= (x x) (x x )

(distribution)

= (x x) (x x )

(commutation)

= (x x) unit

(complementation)

= (x x) (x x )

(complementation)

= (x x ) (x x )

(commutation, twice)

= x (x x )

(distribution)

= x zero

(complementation)

= x

(identity for join)

To establish the first DeMorgan Law, we will use the

Unique Complement Principle for Boolean Algebras:

If x y = unit and x y = zero, then y = x .

4.3 More Results about Boolean Algebras

67

Proof: Assume that (a) x y = unit and (b) x y = zero. Then

y = y unit

(identity for meet)

= y (x x )

(complementation)

= (y x) (y x )

(distribution)

= (x y) (y x )

(commutation)

= zero (y x )

(by assumption (b))

= (x x ) (y x )

(complementation)

= (x x) (x y)

(commutation, twice)

= x

(x y)

(distribution)

= x

unit

(by assumption (a))

=

 

(identity for meet)

x

 

The first DeMorgan Law can now be established as follows:

A. (x y) (x y ) = ((x y) x ) ((x y) y )

(distribution)

= ((y x) x ) ((x y) y )

(commutation)

= (y (x x )) (x (y y ))

(association, twice)

= (y unit) (x unit)

(complementation, twice)

= unit unit

(unit consumption, twice)

= unit

(idempotence)

B.(x y) (x y ) = zero—is left as an exercise.

C.By the Unique Complement Principle, it follows from A and B that (x y) = x y .

The second DeMorgan Law has a dual proof, which is left as an exercise.

Once we have characterized the algebra of truth-values corresponding to a logical system—which is Boolean algebra in the case of classical propositional logic—we can prove things about that system algebraically rather than by reference to truthtables. For example, we know that A ¬A is a tautology of classical propositional logic. But now we can prove this fact algebraically. Under any Boolean algebraic interpretation, the value of A ¬A is x x if the value of A is x, and complementation condition vii sets x x = unit. Thus A ¬A is a tautology under every Boolean algebraic interpretation and therefore a tautology of classical propositional logic. As another example, we have shown that the DeMorgan Laws hold true in every Boolean algebra, and on that basis we may conclude that the formula ¬(A B) of classical propositional logic is equivalent to ¬A ¬B and that ¬(A B) is equivalent to ¬A ¬B. These two equivalences give us the DeMorgan Laws used in Section 2.3 of Chapter 2. Two other equivalences used there, the Distribution equivalences, are the propositional logic counterparts to distribution in Boolean algebras; and Double Negation in that section is the propositional logic version of the Double Negation Law that we have show to hold true in every Boolean algebra.

In Section 4.2 we used the definition P Q = def ¬P Q from classical propositional logic to give us a Boolean algebraic conditional operation satisfying the

68

Alternative Semantics for Truth-Values and Truth-Functions

equation x y = x y. But there is another (equivalent) way to define Boolean algebraic conditional operations, based on the standard lattice-theoretic ordering relation. When considering a lattice as such (i.e., an algebra in which conditions i–iv for Boolean algebras hold), there is a natural ordering relation on elements of the lattice defined as

x y = def x y = x.

We can also use this definition for Boolean algebras since, as we noted earlier, a Boolean algebra is a special type of lattice (namely, a complemented distributed lattice with zero and unit elements). In a Boolean algebra <P(S), , , , S, Ø> (where S is a nonempty set), for example, the ordering relation so defined turns out to be the subset relation —that is, X Y if and only if X Y = X. For the Boolean algebra <{1,0}, max, min, 1–, 1, 0> the ordering relation is simply the numeric ordering relation because here x is less than or equal to y if and only if min(x, y) = x.

In every lattice the relation is reflexive (for all x, x x), antisymmetric (for all x and y, if x y and y x, then x = y), and transitive (for all x, y, and z, if x y and y z, then x z). To show that the relation is reflexive, we note that for any x, x x = x because is defined to be idempotent in every lattice and so x x by the definition of the ordering relation. To show that is antisymmetric we must prove that for any x and y, if x y = x and y x = y, then x = y. This is straightforward, since x y = y x by the requirement of commutativity in every lattice. It is left as an exercise to show that the relation must be transitive in a lattice. We also note that because x y = x if and only if x y = y (see exercises), we also have

x y if and only if x y = y.

We can use the lattice ordering to define conditional operations in Boolean algebras to be operations that satisfy

x y = unit if and only if x y.

In the algebra of classical truth-values, for example, this yields exactly the logical conditional operation. Consider the classical truth-values under their lattice ordering. Because the ordering is reflexive we have T T and F F. Moreover, because zero unit = unit, we will always have zero unit and so F T. However, the converse (unit zero) does not hold when unit and zero are distinct. Given this ordering, the truth-value of a conditional P Q in classical logic is T if and only if V(P) V(Q).

Before closing, we note that in addition to characterizating the semantics for classical propositional logic there is another important and related way that Boolean algebras are used to study classical propositional logic. For any logical system there is a special type of algebra called a Lindenbaum algebra, constructed from equivalence classes of formulas in the system, and the Lindenbaum algebras for