Lecture 5: Supplementary Note on Huntintong’s Postulates

Instructor: Yong Kim (Section 1)

Motivation: Why should we care about axioms, postulates and theorems? Axioms and Postulates are given facts we don’t need to prove, but theorems are proven using axiom and postulates. If we don’t know which ones are postulates or theorems, how can we figure out how to prove if given expressions are identical? Knowing exactly which are postulates and theorems should give us better understanding on how we can find alternative expressions of a given expression.

Basic Definitions of Boolean Algebra

Boolean Algebra

  • introduced by George Boole in 1854
  • a set of elements: E = {0, 1}
  • a set of operators: O = {+, , ‘}

binary operator {+,}: works on two operands

unary operator {¯}: works on one operand

  • a number of axioms or postulates: (assumptions – do not need to be proved)
  • axiom: a proposition that is assumed without proof for the sake of studying its consequences
  • postulate: assume without proof (generally ‘because its obvious)
  • a number of theorems (can be proven from the postulates)

Common postulates used to formulate algebraic structures

1. Closure

A set S is closed with respect to a binary operator if, for every pair of elements of S, the binary operator specifies a rule for obtaining a unique element of S.

example:

The set of natural numbers N = {1, 2, 3, 4, …} is closed with respect to the binary operator plus (+) by the rules of arithmetic addition, since a, bNwe obtain a unique cN by the operation a + b = c.

The set of natural numbers is not closed with respect to the binary operator minus (-) by the rules of arithmetic subtraction because 2 – 3 = -1 and 2, 3 N, while (-1) N.

2. Associative law

A binary operator * on a set S is said to be associative whenever

(x * y) * z = x * (y * z)for all x, y, zS

3. Commutative law

A binary operator * on a set S is said to be commutative whenever

x * y = y * xfor all x, yS

4. Identity element

A set S is said to have an identity element with respect to a binary operation * on S if there exists an element eS with the property

e * x = x * e = xfor any xS

example:

The element 0 is an identity element with respect to operation + on the set of integers I = {…, -3, -2, -1, 0, 1, 2, 3, …} since

x + 0 = 0 + xfor any xI

5. Inverse

A set S having the identity element e with respect to a binary operator * is said to have an inverse whenever, for every xS, there exists an element yS such that

x * y = e

6. Distributive law

If * and  are two binary operators on a set S, * is said to be distributive over  whenever

x * (yz) = (x * y)  (x * z)

Ordinary algebra
  • The binary operator + defines addition.
  • The additive identity is 0.
  • The additive inverse defines subtraction.
  • The binary operator  defines multiplication.
  • The multiplicative identity is 1.
  • The multiplicative inverse of a = 1/a defines division, i.e., a 1/a = 1.
  • The only distributive law applicable is that of  over +:

a (b + c) = (ab) + (ac)

Huntington Postulates

Our book mixes up postulates and theorems in Mano & Kime, p 33, Table 2-3 and call everything identities. It may be simple to put everything as identities, but Huntington has proposed several important postulates and everything else (mainly theorems) can be proven using these postulates.

Let’s us see if we can show if Boolean algebra satisfies common properties required by algebra.

1. Closure

Closure is obvious from the tables since the result of each operation is either 1 or 0 and 0, 1  B.

2. Identity elements (from the tables)

(a) 0 + 0 = 00 + 1 = 1 + 0 = 1

(b) 1  1 = 11  0 = 0  1 = 0

3. Commutative law

Commutivity is obvious from the symmetry of the binary operator tables.

4. Distributive law x (y + z) = (xy) + (xz)

x / y / z / y + z / x(y + z) / xy / Xz / (xy) + (xz)
0 / 0 / 0 / 0 / 0 / 0 / 0 / 0
0 / 0 / 1 / 1 / 0 / 0 / 0 / 0
0 / 1 / 0 / 1 / 0 / 0 / 0 / 0
0 / 1 / 1 / 1 / 0 / 0 / 0 / 0
1 / 0 / 0 / 0 / 0 / 0 / 0 / 0
1 / 0 / 1 / 1 / 1 / 0 / 1 / 1
1 / 1 / 0 / 1 / 1 / 1 / 0 / 1
1 / 1 / 1 / 1 / 1 / 1 / 1 / 1

5. Complement

(a) x + x’ = 1, since 0 + 0’ = 0 + 1 = 1 and 1 + 1’ = 1 + 0 = 1.

(b) xx’ = 0, since 0  0’ = 0  1 = 0 and 1  1’ = 1  0 = 0

which verifies postulate 5.

6. Postulate 6

Postulate 6 is satisfied because the two-valued Boolean algebra has two distinct elements, 1 and 0, with 1  0.

Now, we can summarize all the above and figure out which ones are corresponding identities from our book, Mano & Kime, p 33, Table 2-3.

1.(a) Closure with respect to the operator +.

(b) Closure with respect to the operator .

2.(a) An identity element with respect to +, designated by 0:(Identities 1-4 from Mano & Kime, p 33, Table 2-3)

x + 0 = 0 + x = x

(b) An identity element with respect to , designated by 1:

x 1 = 1 x = x

3.(a) Commutative with respect to +:(Identities 11-12)

x + y = y + x

(b) Commutative with respect to :

xy = yx

4.(a)  is distributive over +:(Identities15-16)

x (y + z) = (xy) + (xz)

(b) + is distributive over :

x + (yz) = (x + y)  (x + z)

5.For every element xB, there exists an element x’ B such that(Identities 7-8)

(a) x + x’ = 1 and

(b) xx’ = 0

x’ is called the complement of x

6.There exists at least two elements x, yB such that xy.

Now, here is the original form of Huntington’s Postulates and Theorems.

Postulates and Theorems of Boolean Algebra
Postulate 2 / (a) x + 0 = x / (b) x 1 = x
Postulate 5 / (a) x + x’ = 1 / (b) xx’ = 0
Theorem 1 / (a) x + x = x / (b) xx = x
Theorem 2 / (a) x + 1 = 1 / (b) x 0 = 0
Theorem 3,
involution / (x’)’ = x
Postulate 3, commutative / (a) x + y = y + x / (b) xy = yx
Theorem 4,
associative / (a) x + (y + z)
= (x + y) + z / (b) x(yz) = (xy)z
Postulate 4, distributive / (a) x(y + z) = xy + xz / (b) x + yz
= (x + y)(x + z)
Theorem 5,
DeMorgan / (a) (x + y)’ = x’y’ / (b) (xy)’ = x’ + y’
Theorem 6,
absorption / (a) x + xy = x / (b) x(x + y) = x

We can prove his theorems using his postulates,

Say, proof of Theorem 1(b):xx = x

xx = xx + 0by Postulate 2(a)

= xx + xx’by Postulate 5(b)

= x(x + x’)by Postulate 4(a)

= x 1by Postulate 5(a)

= xby Postulate 2(b)

Similarly, you should try to prove all his theorems on your own.