4.2 Boolean Algebras and Classical Logic

The notation min{Iv (P): v is an x-variant of v} means: the minimum value that P has on any x-variant of v, and similarly for max{Iv (P): v is an x-variant of v}. Thus a universally quantified formula has the value 1 only if the formula being quantified over has the value 1 for every value of x, and an existentially quantified formula will have the value 1 if the formula being quantified over has the value 1 for at least one value of x. Note that the use of the maximum and minimum functions in clauses 7 and 8 reflects our understanding of the two quantifiers, respectively, in terms of conjunction (min) and disjunction (max).

4.2 Boolean Algebras and Classical Logic

We will be interested in numeric values in the following chapters, but we will also be interested in more abstract characterizations of the semantic structures for our logics, particularly in the case of fuzzy logics. The semantics for classical logic can be studied abstractly using Boolean algebras.

To motivate this abstraction, we point out it’s possible to use yet other pairs of values for classical propositional logic in place of true and false or 1 and 0 as long as we define operations for the propositional connectives that preserve the structure of the classical values. As another example, we could use the set {5} in place of true and the empty set Ø in place of false and then define the values for complex formulas of classical propositional logic as follows, based an assignment V of one or the other of these values to each atomic formula:

1.V(¬P) = {5} – V(P)

where − is set-theoretic difference, that is, X−Y is the set consisting of all the members of X that are not members of Y; so that V(¬P) is Ø if V(P) is {5}, and V(¬P) is {5} when V(P) is Ø.

2. V(P Q) = V(P) V(Q),

where is set-theoretic intersection: X Y is the set consisting of all items that are members of both X and Y. Thus V(P Q) = {5} if both V(P) and V(Q) are {5}, and Ø otherwise.

3. V(P Q) = V(P) V(Q))

where is set-theoretic union: X Y is the set consisting of all items that are members of X or of Y or both. So V(P Q) is {5} if V(P) or V(Q) is, and Ø otherwise.

4.V(P → Q) = {5} if V(P) V(Q), and Ø otherwise

where is set-theoretic inclusion: X Y if (and only if) every member of X is also a member of Y—so the only case in which V(P → Q) is Ø occurs when V(P) is {5} and V(Q) is Ø.

5. V(P ↔ Q) = (left as an exercise).

The structure imposed on truth-values by the semantic operations of classical propositional logic—a structure mirrored in the interpretations based on 1 and 0 and on {5} and Ø—yields a Boolean algebra.1 A Boolean algebra <B, , ∩, , unit, zero> consists of a set B (the domain of the Boolean algebra) that contains at least two elements designated as the unit and zero elements, binary operations and ∩ (respectively called join and meet),2 and a unary operation (called complementation), such that the following conditions are satisfied for all members x, y, z of B:

Condition i stipulates that the meet and join operations are commutative; condition ii stipulates that they are associative; and condition iii stipulates that they are idempotent. Condition iv specifies absorption laws, while condition v stipulates that each of the two binary operations distributes over the other. Condition vi stipulates that the zero element and unit elements are, respectively, identity elements for the join operation and the meet operation ∩, and condition vii specifies complementation laws.

We’ll describe the Boolean algebra based on the numeric values 1 and 0. The domain B is the set {1, 0}, where 1 is the unit member and 0 is the zero member. The maximum and minimum operations—which define disjunction and conjunction— serve as the algebra’s meet ( ) and join (∩) operations; and the negation operation 1−,which produces 1−xwhen applied to x, serves as complementation. It is straightforward to verify that the seven conditions on a Boolean algebra’s operations are met for the structure <{1,0}, max, min, 1−, 1, 0>, that is, for all x, y, z {1, 0}:

i.Commutation: max (x, y) = max (y, x), and min (x, y) = min (y, x) Proof: Obviously true.

ii.Association: max (x, max (y, z)) = max (max (x, y), z), and min (x, min (y, z)) = min (min (x, y), z)

Proof: Obviously true.

1Our definition of Boolean algebras and of lattices (introduced later) follows MacLane and Birkhoff (1999). Alternative (equivalent) definitions of Boolean algebras appear in various places in the literature. These algebras are named after the mathematician George Boole, who first developed them to study logic.

2The join and meet symbols are standardly used to denote set-theoretic union and intersection— but in the context of Boolean algebras they are used to denote any operations, set-theoretic or otherwise, that meet the specified conditions.

iii.Idempotence: max (x, x) = x, and min (x, x) = x. Proof: Obviously true.

iv.Absorption: max (x, min (x, y)) = x, and min (x, max (x, y)) = x.

Proof of first equation: For any x, y {1, 0} there are two possibilities: either x ≥ y, or x < y. If x ≥ y, then max (x, min (x, y)) = max (x, y) = x. If x < y, then max (x, min (x, y)) = max (x, x) = x. Either way, the equation holds.

Proof of second equation: Left as an exercise.

v.Distribution: max (x, min (y, z)) = min (max (x, y), max (x, z)), and min (x, max (y, z)) = max (min (x, y), min (x, z)).

Proof of first equation: We’ll consider the different orderings that x, y, and

zcan have, as three cases:

a.If x ≥ y and x ≥ z, then max (x, min (y, z)) = x = min (x, x)

=min (max (x, y), max (x, z)) because max (x, y) = max (x, z) = x

b.If y ≥ x and z ≥ x, then max (x, min (y, z)) = min (y, z) = min (max (x, y), max (x, z)) because max (x, y) = y and max (x, z) = z

c.If either y ≥ x ≥ z or z ≥ x ≥ y, then max (x, min (y, z)) = x because

x≥ min (y, z), and

if y ≥ x ≥ z then min (max (x, y), max (x, z)) = min (y, x) = x while if z ≥ x ≥ y then min (max (x, y), max (x, z)) = min (x, z) = x

Proof of second equation: Left as an exercise.

vi.Identity for join and meet: max (x, 0) = x, and min (x, 1) = x.

Proof: Obviously true.

vii.Complementation: max (x, 1 – x) = 1, and min (x, 1 – x) = 0.

Proof of first equation: If x = 1, then max (x, 1 – x) = max (1, 0) = 1. If x = 0, then max (x, 1 – x) = max (0, 1) = 1.

Proof of second equation: Left as an exercise.

We can similarly prove that either the values T and F or the values {5} and Ø taken, respectively, as unit and zero elements, along with the corresponding operations defining disjunction, conjunction, and negation, form Boolean algebras. When we interpret formulas of propositional logic based on a two-valued Boolean algebra by assigning either unit or zero to each atomic formula and using the algebra’s join, meet, and complement operations to define the, respective, values of disjunctions, conjunctions, and negations, we call this an algebraic interpretation based on that Boolean algebra and the set of all such interpretations a semantics based on the Boolean algebra. We will say that a formula of propositional logic is a tautology of a Boolean algebra (a BA-tautology) if the formula evaluates to unit under every algebraic interpretation based on that algebra. For every two-valued Boolean algebra we obtain the same set of tautologies for propositional logic (and the same entailments, etc.) owing to the following result:

Result 4.1: Every two-valued Boolean algebra BA = <{unit, zero}, , ∩, , unit, zero> generates the following truth-tables for assignments of unit or zero to each

atomic formula of propositional logic when , ∩, and , respectively, define the disjunction, conjunction, and negation operations:

Proof: This follows from the four Boolean algebra conditions i. x y = y x, and x ∩ y = y ∩ x

iii. x x = x, and x ∩ x = x

vi.x zero = x, and x ∩ unit = x

vii.x x = unit, and x ∩ x = zero

Consider the table for conjunction. By condition iii, unit ∩ unit = unit and zero ∩ zero = zero. That gives us the first and fourth rows. By condition vi, zero ∩ unit = zero, which gives us the third row. By condition i, unit ∩ zero = zero ∩ unit and so unit ∩ zero = zero as well, which gives us the second row. The reader will be asked in the exercises to verify that the four conditions also generate the displayed tables for disjunction and negation.

It turns out that conditions ii, iv, and v defining Boolean algebras (the conditions that were not used in the proof of Result 4.1) can be derived from the remaining four conditions when the algebra is two-valued. This will be considerably easier to show after we prove some additional equations that hold in Boolean logics in Section 4.3, so we will defer proof to the exercises for that section.

To round out the Boolean operations corresponding to connectives of propositional logic, we’ll use the definitions P → Q = def ¬P Q and P ↔ Q = def (P → Q) (Q → P). As a consequence the Boolean algebraic operations and corresponding to the conditional and biconditional satisfy the equations x y = x y and x y = (x y) ∩ (y x). With these definitions the following is a direct consequence of Result 4.1:

Result 4.2: For any two-valued Boolean algebra BA, the set of formulas of propositional logic that are BA-tautologies is exactly the set of classical tautologies under the standard semantics based on T and F.

Analogously, entailments of propositional logic under any two-valued Boolean semantics—where a set of formulas entails a formula P if P evaluates to the