Table Of ContentLectures in
Discrete Mathematics
Edward A. Bender
S. Gill Williamson
Edward A. Bender & S. Gill Williamson
2010
Arithmetic, Logic and Numbers
With and Introduction to Cryptography
Unit BF: Boolean Functions and Computer Arithmetic
Edward A. Bender
S. Gill Williamson
Unit BF
Boolean Functions and Computer Arithmetic
Section 1: Boolean Functions
We recall the concept of a function and some of the terminology.
Definition 1 (Function) If A and B are sets, a function from A to B is a rule that
tells us how to find a unique b∈ B for each a∈ A. We write f(a) = b and say that f maps
a to b. We also say the value of f at a is b.
We write f : A → B to indicate that f is a function from A to B. We call the set A
the domain of f and the set B the range or, equivalently, codomain of f.
To specify a function completely you must give its domain, range and rule.
In this section, we’ll study a special class of functions called “boolean functions.”
General properties of functions are studied in Unit SF.
A Boolean function is a function f from the Cartesian product ×n{0,1} to {0,1}.
Alternatively, we write f : ×n{0,1} → {0,1}. The set ×n{0,1}, by definition, the set of
all n-tuples (x1,···,xn) where each xi is either 0 or 1, is called the domain of f. The set
{0,1} is called the codomain (or, sometimes, range) of f. The Cartesian product ×n{0,1}
is also written {0,1}n. This corresponds to writing the product of n copies of y as yn.
Example 1 (Tabular representation of Boolean functions) One way to represent a
function whose domain is finite is with a table. Each element x of the domain has a row
of the table listing the domain element x and the corresponding function value f(x). For
example, the two tables
p q r g
0 0 0 1
p q f
0 0 1 1
0 0 0 0 1 0 0
0 1 1 0 1 1 0
1 0 0 1 0 0 0
1 1 1 1 0 1 0
1 1 0 1
1 1 1 1
define Boolean functions f : ×2{0,1} → {0,1} and g : ×3{0,1} → {0,1}. In the case of
f, the first two values of each row represent the argument of f and the third entry in the
same row represents the value of f at that argument. We have f(0,0) = 0, f(0,1) = 1,
f(1,0) = 0, and f(1,1) = 1. For g, the first three values of each row represent an element
of ×3{0,1} (a triple (p,q,r) of 0’s or 1’s) and the fourth entry of the row represents the
value of g(p,q,r). Thus, g(0,0,0) = 1, g(0,0,1)= 1, g(0,1,0)= 0, etc.
BF-1
Boolean Functions and Computer Arithmetic
Noticethatwehaveusedp,q andrforvariablesinsteadofx,y andz,whichareusually
the choice for variable names in algebra. We can choose any names we please for variables.
Names such as p, q and r are commonly used in the study of Boolean functions.
The tabular form of a Boolean function is often called a truth table because of the
connection between Boolean functions and logic, which we will study later.
Example 2 (The number of Boolean functions) How many Boolean functions are
there with domain ×3{0,1}? More generally, how many Boolean functions are there with
domain ×n{0,1}? We do this in two steps.
First, how many elements are there in the domain ×n{0,1}? We use the notation
|S| to denote the number of elements of S (also called the cardinality of S). Thus we are
asking for the value of |×n{0,1}|. We can select an element of the domain by making n
choices x1,x2,...,xn where xi ∈ {0,1} for i = 1,2,...,n. Thus there are two choices for
x and two choices for x and so on. The total number of elements in the domain is thus
1 2
2×2×···×2 = 2n. In other notation, |×n{0,1}| = |{0,1}|n. If you have trouble seeing
why this is true, the footnote may help.1
Second, we must construct the Boolean function. In constructing a Boolean function
h : ×3{0,1} → {0,1}, there are two possible choices that we could assign to h(p,q,r) for
each (p,q,r) ∈ ×3{0,1}. We saw in the previous paragraph that there are 23 elements
(p,q,r) in ×3{0,1}. Thus the total number of Boolean functions h : ×3{0,1} → {0,1} is
2(23). In general, the number of Boolean functions h :×n{0,1} →{0,1} is 2(2n).
In this section we are concerned about a particular way of representing Boolean func-
tions in terms of certain more “primitive” representations. An example that is familiar
from high school mathematics is the representation of polynomial and rational functions,
such as 1+x, 1+x2, (2+x2−x3)/(1+x2), starting with the constant functions c and the
identity function x. All of these functions are created by adding, multiplying, subtracting,
and/or dividing the simple starting functions.
Example 3 (The simplest Boolean functions: constants, identity, not) The
simplest functions are the constant functions. Since we have only two constants, 0 and 1,
there are two constant Boolean functions.
The next simplest Boolean functions are those that depend only on a single variable,
butarenotconstant. Sincethereare221 = 22 = 4one-variableBooleanfunctionsandsince
there are two constants, there are 4−2 = 2 nonconstant, one-variable Boolean functions.
One is the identity function f(p) = p for all p ∈ {0,1}. The other one-variable function is
“not p.” For the moment, let’s call this function n(p). The definition is simple: n(p) = 0
if p = 1, n(p)= 1 if p = 0. This function is usually denoted by ∼p rather than n(p). Thus,
1 How many elements are there in the Cartesian product S×T of a set with s elements
and a set with t? Imagine an s by t array. Can you see why each entry in the array can
be thought of as an element of S ×T? If so, we are almost done. We have shown that
|S ×T| = |S|·|T|. Apply this over and over again to {0,1}×{0,1}×···×{0,1}.
BF-2
Section 1: Boolean Functions
we would write ∼0 = 1 and ∼1 = 0. The symbol “∼” is called the “unary operator not.”
Unary means it is a function that has one variable. The minus sign in ordinary arithmetic
is a familiar example of a unary operator.
Example 4 (Two-variable Boolean functions: and, or, exculsive or) We know
thereare222 = 24 = 16two-variableBooleanfunctionsf(p,q). Fromthepreviousexample,
2 of these are constant, 2 of them are not constant and depend only on p, and 2 of them
are not constant and depend only on q. This leaves 16−6 = 10 functions that depend on
both variables. We certainly don’t want to give names to all of them! In this example, we
define three commonly-used, two-variable Boolean functions.
Thefirstfunctionwedefineis“pandq.” Again,forthemoment,let’scallthisfunction
(a function of two Boolean variables) a(p,q). By definition, a(p,q) = 0 unless both p = 1
and q = 1, in which case a(p,q) = 1. The function a(p,q) is denoted by p∧q. We write
0∧0 = 0, 0∧1 = 0, 1∧0 = 0, and 1∧1 = 1.
Thenextfunctionis “p orq.” Again, for themoment,let callthis function(a function
of two Boolean variables) o(p,q). By definition, o(p,q) = 1 unless both p = 0 and q = 0,
in which case o(p,q) = 0. The function o(p,q) is denoted by p∨q. We write 0∨0 = 0,
0∨1 = 1, 1∨0 = 1, and 1∨1 = 1.
The last function is the exclusive or function. It is written xor(p,q) or p ⊕ q. By
definition p⊕q = 0 if p = q and p⊕q = 1 if p 6= q.
Here are our functions in tabular form:
p q p∧q p q p∨q p q p⊕q
0 0 0 0 0 0 0 0 0
0 1 0 0 1 1 0 1 1
1 0 0 1 0 1 1 0 1
1 1 1 1 1 1 1 1 0
The symbols ∨, ∧, and ⊕ are called “binary operators” because they involve two variables
(like addition and multiplication in ordinary arithmetic).
Example 5 (Boolean functions and logic) People often call Boolean variables such
as p and q statement variables. Why? Because Boolean functions are related to logic. In
logic we think of p and q as statements, 0 as “false,” and 1 as “true.”
Suppose p stands for “I have classes tomorrow” and q stands for “I will stay home
tomorrow.” Let’s look at our basic Boolean functions.
• ∼p stands for “not (I have classes tomorrow),” which can be written in more normal
English as “I do not have classes tomorrow.” As mentioned earlier, 0 is thought of as
“false” and 1 is thought of as “true.” According to our definition of the function ∼, if
p is true, then ∼p is false. This is also true about our statements: If the statement “I
have classes tomorrow” is true, then the statement “I do not have classes tomorrow”
is false.
• p∧qstandsforthestatement,“IhaveclassestomorrowandIwillstayhometomorrow.”
You should verify that the definition of p∧q agrees with our usual interpretation of
BF-3
Boolean Functions and Computer Arithmetic
“and”: p∧q is true if and only if both p and q are true. The statement p∧q is also
read “p but q”, especially if q is surprising as in “I have classes tomorrow, but I will
stay home tomorrow.”
• p∨ q stands for the statement, “Either I have classes tomorrow or I will stay home
tomorrow.” Unfortunately “or” is ambiguous in English. If I have classes and also
stay home, some people may think the statement is not true. Others may think that
it is true. Our definition of or is not ambiguous: p∨q is true if either p or q or both
are true.
• p ⊕ q stands for the statement “Either I have classes tomorrow or I will stay home
tomorrow, but not both.” Why is this? The value of p⊕q is 1 (true) if and only if one
of p and q is 1 (true) and the other is 0 (false).
When converting ordinary language into symbolic form, it is important to be aware
of ambiguities so that the statements can be converted correctly. What we have been
discussing is called propositional logic. We will explore the connection between Boolean
functions and logic in the unit on logic.
Because of the close connection with logic, the tabular form of a Boolean function is
often called a truth table.
Why are the Boolean functions represented by ∼, ∧, ∨ and ⊕ important? In the
previousexample, wehaveseenahintoftheirimportanceinlogic. Theyarealsoimportant
because they can be used to define more complex Boolean functions, in the same way that
the basic operations of arithmetic can be used to define more complex algebraic functions.
Example 6 (Defining more complex Boolean functions) We could try to define a
Boolean function using ∼, ∧, and ∨ by stating that
f(p,q) = p∧∼q∨∼p∧q. (ambiguous)
The problem with this “definition” is that it is not clear what the order of application of
these operators should be. It is conventional to give top priority to the ∼ operator. Thus
the expression used to define f can be clarified a bit: f(p,q) = p∧ (∼q)∨(∼p)∧ q. In
addition,theorderin which the∧ and∨ areperformedmust bespecifiedby groupingthem
in the definition of the function f. Thus, as an example, we could group them
f(p,q) = (p∧(∼q))∨((∼p)∧q). (not ambiguous)
Now thefunctionf is clearlydefined. You shouldmake surethatyou use enoughparenthe-
ses. Using toofewmayresultin anambiguousfunctiondefinition. Using morethanneeded
does not changea function. For example, the usage of ∼ is definedby the precedencerules,
so we could just write
f(p,q) = (p∧∼q)∨(∼p∧q). (still not ambiguous)
If you’re unsure about precedence rules, be safe and use extra parentheses!
BF-4
Section 1: Boolean Functions
Using the formula for f we can compute values and give f in tabular form:
p q f
0 0 0
0 1 1
1 0 1
1 1 0
This is the exclusive or function. We have proved p⊕q = (p∧∼q)∨(∼p∧q).
Inthepreviousexample, wefoundthatp⊕q couldbewrittenasanequivalentBoolean
function using ∼, ∧ and ∨. This is a particular example of a much more general result:
Theorem 1 (Representing functions) Suppose n > 0 and f : ×n{0,1} → {0,1}.
There is a function g using only ∼, ∨ and ∧ that is equal to f. In fact, we can use just ∼
and ∧ or, if we prefer, we can use just ∼ and ∨.
Proof: We’ll illustrate how to do this with the function
p q r f
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1
Look at the rows in the table where the f = 1. The first such is (p,q,r) = (0,0,1). Note
that (∼p)∧((∼q)∧r) equals 1 when (p,q,r) = (0,0,1) and equals 0 otherwise.2 Similarly,
for the rows (p,q,r) = (0,1,0) and (p,q,r) = (1,1,1), we have (∼p) ∧ (q ∧ (∼r)) and
p∧(q∧r). You should be able to see why the function
g(p,q,r) = (∼p)∧((∼q)∧r) ∨ (∼p)∧(q∧(∼r)) ∨ p∧(q∧r)
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
is equal to f.
It should be clear how to do this in general: Construct an appropriate “anding” for
each row of the table where the function equals 1. Then “or” all these andings together.
This proves the first claim.
In this paragraph we take a break to discuss some terminology. An expression of this
sort, namelyan “or” of “ands”of variables and their negations,is called disjunctive normal
form. Wheredid“disjunctive”comefrom? An“or”issometimescalledadisjunction. Ifwe
2 We don’t need all these parentheses. The value of ∼p∧∼q∧r is unambiguous. We’re
just playing it safe!
BF-5
Boolean Functions and Computer Arithmetic
replace the roles of “and” and “or,” we obtain conjunctive normal form, so called because
“and” is called a conjunction. End of terminology discussion and back to the proof.
One might object that we haven’t taken care of the case when the function is a con-
tradiction since then there are no rows where the function is 1. This is taken care of by 0
or, if you prefer, by p∧∼p.
How can we get rid of ∨? We claim that
P ∨Q∨R∨···∨T and ∼ (∼P)∧(∼Q)∧(∼R)∧···∧(∼T)
(cid:16) (cid:17)
are equal. Why? The only way the first expression can be 0 is for all of P,Q,...,T to
be 0. The only way the second expression can be 0 is for the expression inside the large
parentheses to be 1. The only way this can happen is for all of ∼P,∼Q,...,∼T to be 1 —
which is the same as all of P,Q,...,T being 0. Applying this to g with
P = (∼p)∧((∼q)∧r), Q = (∼p)∧(q∧(∼r)), R = p∧(q∧r),
and none of the terms ...,T, we have the equivalent function
∼ ∼ (∼p)∧((∼q)∧r) ∧∼ (∼p)∧(q∧(∼r)) ∧∼ p∧(q∧r) .
(cid:18) (cid:19)
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
You should see that this works in general.
What about getting rid of ∧ instead of ∨? A similar trick can be used, working from
the “inside” of the formula instead of from the “outside.” We leave it to your inventiveness
to find the trick.
In the last half of the previous proof, we used a trick that let us replace ∨ with ∧. In
fact, this trick is used often enough that we should call it a rule.3 It is very useful to have
a catalog of simple rules to help in deciding whether or not functions are equal, without
having to always construct tables of functions. Here is such a catalog. In each case the
standard name of the rule is given first, followed by the rules as applied first to ∧ and then
to ∨.
Theorem 2 (Algebraic rules for Boolean functions) Each rule states that two
different-looking Boolean functions are equal. That is, they look different but have the
same table.
Associative Rules: (p∧q)∧r = p∧(q∧r) (p∨q)∨r = p∨(q∨r)
Distributive Rules: p∧(q∨r)= (p∧q)∨(p∧r) p∨(q∧r) = (p∨q)∧(p∨r)
Idempotent Rules: p∧p = p p∨p = p
Double Negation: ∼∼p = p
DeMorgan’s Rules: ∼(p∧q) = ∼p∨∼q ∼(p∨q) = ∼p∧∼q
Commutative Rules: p∧q = q∧p p∨q = q∨p
Absorption Rules: p∨(p∧q) = p p∧(p∨q) = p
Bound Rules: p∧0 = 0 p∧1 = p p∨1 = 1 p∨0 = p
Negation Rules: p∧(∼p) = 0 p∨(∼p)= 1
3 It is sometimes said that a rule or method is a trick that is used more than once.
BF-6
Section 1: Boolean Functions
These rules are “algebraic” rules for working with ∧, ∨, and ∼. You should memorize
them as you use them. They are used just like rules in ordinary algebra: whenever you
see an expression on one side of the equal sign, you can replace it by the expression on the
other side. Each of the rules can be proved by constructingtables for the functions on each
side of the equal sign and verifying that those tables give the same function values.
Truth tables are similar to the tabular method for proving set identities (see Section 1
of Unit SF). The algebraic rules for Boolean functions are almost identical to the rules
for sets in Section 1 of Unit SF. When two apparently very different situations (sets and
Boolean functions in this case) are similar, one should look for an explanation. We provide
an explanation in Unit Lo.
Oneusefulconsequenceofthisconnectionis thatwecanuse“Venndiagrams”toprove
identities for Boolean functions. (If you are not familiar with Venn diagrams, read the first
four pages of Unit SF.) How does this work? Think of the variables p, q and so on as sets.
Then make the translations:
∧ to ∩, ∨ to ∪, ∼ to complement.
The universal set is 1 and the empty set is 0. For example, you can prove the first of
c c c
DeMorgan’s Rules for Boolean functions, by showing that (P ∩Q) and P ∪Q give the
same regions in the Venn diagram for two sets P and Q. Try it.
Example 7 (Manipulating functions) We want to simplify the function
(∼(p∧∼q))∧(p∨q).
Here are our calculations:
(∼(p∧∼q))∧(p∨q) = (∼p∨∼∼q)∧(p∨q) (DeMorgan’s rule)
= (∼p∨q)∧(p∨q) (double negation)
= ((∼p∨q)∧p)∨((∼p∨q)∧q) (distributive rule)
= (p∧(∼p∨q))∨(q∧(q∨∼p)) (commutative rule 3 times)
= (p∧(∼p∨q))∨q (absorbtion rule)
= ((p∧∼p)∨(p∧q))∨q (distributive rule)
= (0∨(p∧q))∨q (negation rule)
= (p∧q)∨q (bound rule)
= q (commutative and absorbtion rules)
Except for the last step, we gave each step in detail. In actual calculations, you can
combine steps as we did in the last step. How you decide to manipulate things can make
a big difference. For example,
(∼(p∧∼q))∧(p∨q)= (∼p∨∼∼q)∧(p∨q) (DeMorgan’s rule)
= (∼p∨q)∧(p∨q) (double negation)
= (q∨∼p)∧(q∨p) (commutative rule)
= q∨(∼p∧p) (distributive rule)
= q∨0 (negation rule)
= q (bound rule)
BF-7
Boolean Functions and Computer Arithmetic
The same thing happens in algebra; however, you are more likely to do things the shorter
way in algebra because you are more familiar with those manipulations. There is a trade
off between taking time to try finding a shorter way and simply going ahead. This is
a problem faced by designers of “symbolic manipulation” packages such as Maple(cid:13)R and
Mathematica(cid:13)R.
Exercises for Section 1
1.1. Let f = “she is out of work” and s = “she is spending more.” Write the following
statements in symbolic form:
(a) She is out of work but she is spending more.
(b) Neither is she out of work nor is she spending more.
1.2. Letr =“sheregisteredtovote”andv =“shevoted.” Writethefollowingstatement
in symbolic form: She registered to vote but she did not vote.
1.3. Make a truth table for ∼ (p∧q)∨∼(p∨q) .
(cid:0) (cid:1)
1.4. Make a truth table for ∼p∧(q∨∼r).
1.5. Make a truth table for p∨(∼p∨q) ∧∼(q∧∼r).
(cid:0) (cid:1)
1.6. Using DeMorgan’s rule, state the negation of the statement: “Mary is a musician
and she plays chess.”
1.7. Using DeMorgan’s rule, state the negation of the statement: “The car is out of gas
or the fuel line is plugged.”
1.8. Show that p∨(p∧q) = p follows from the idempotent rule, distributive rule, and
the absorption rule p∧(p∨q) = p.
1.9. Is the function (p∧q)∨r equal to the function p∧(q∨r)?
1.10. Is the function (p∨q)∨(p∧r) equal to the function (p∨q)∧r?
1.11. Is the function (∼p∨q)∧(p∨∼r) ∧(∼p∨∼q) equal to the function ∼(p∨r)?
(cid:0) (cid:1)
1.12. Is the function (r∨p)∧ ∼r∨(p∧q) ∧(r∨q) equal to the function p∧q?
(cid:0) (cid:1)
1.13. Is the function ∼(p∨∼q)∨(∼p∧∼q) equal to the function ∼p?
1.14. Is the function ∼ (∼p∧q)∨(∼p∧∼q) ∨(p∧q) equal to the function ∼p?
(cid:0) (cid:1)
1.15. Is the function p∧ ∼(∼p∨q) ∨(p∧q) equal to the function p∨q?
(cid:16) (cid:17)
(cid:0) (cid:1)
BF-8
