# Lecture 5: Supplementary Note on Huntintong’s Postulates

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10 Οκτ 2013 (πριν από 4 χρόνια και 9 μήνες)

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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 axi
om
and postulates. If we don’
t know which ones are

postulate
s

or theorem
s
,
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
alt
ernative 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 operat
or
{
¯
}:

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 ob
vious)

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 specifi
es 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

N

we 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
associ
ative

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

*
x

for all
x
,
y

S

4. Identity element

A set
S

is said to have an
identity element

w
ith respect to a binary
operation * on
S

if there exists an element
e

S

with the property

e

*
x

=
x

*
e = x

for 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 +
x

for 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.

0
.

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 Postul
ates

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 theorem
s) 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 = 0

0 + 1 = 1 + 0 = 1

(b)
1

1 = 1

1

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

X
z

(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 ab
ove 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 resp
ect to

:

x

y

=
y

x

4.

(a)

is distributive over +:

(
Identities

15
-
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

Pos
tulate 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

+ 0

by Postulate 2(a)

=
xx

+
xx

by Postulate 5(b)

=
x
(
x

+
x
’)

by Postulate 4(a)

=
x

1

by Postulate 5(a)

=
x

by Pos
tulate 2(b)

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