Discrete Mathematics CS 2610

February 17

2

Equal Boolean Functions

Two Boolean functions F and G of degree n are equal iff for all (x1,..xn) Bn, F (x1,..xn) = G (x1,..xn)

Example: F(x,y,z) = x(y+z), G(x,y,z) = xy + xz, and F=G (recall the “truth” table from an earlier slide)Also, note the distributive property:

x(y+z) = xy + xz via the distributive law

3

Boolean Functions

Two Boolean expressions e1 and e2 that represent the exact same function F are called equivalent

F(x1,x2,x3)

1110

1010

0100

1000

x3x2x1

0001

0101

0011

1111

F(x1,x2,x3) = x1(x2+x3)+x1x2x3

F(x1,x2,x3) = x1x2+x1x3+x1x2x3

4

Boolean Functions

More equivalent Boolean expressions: (x + y)z = xyz + xyz + xyz (x + y)z = xz + yz distributive = x1z + 1yz identity = x(y + y)z + (x + x)yz unit = xyz + xyz + xyz + xyz distributive = xyz + xyz + xyz idempotent

We’ve expanded the initial expression into its sum of products form.

5

Boolean Functions

More equivalent Boolean expressions: xy + z = ?? = xy1 + 11z identity = xy(z + z) + (x + x)1z unit = xyz + xyz + x1z + x1z distributive = xyz + xyz + x(y + y)z + x(y + y)z unit = xyz + xyz + xyz + xyz + xyz + xyz

distributive = xyz + xyz + xyz + xyz + xyz

idempotent

6

Representing Boolean Functions

How to construct a Boolean expression that represents a Boolean Function ?

1110

1010

0100

0000

zyx

0001

1101

0011

1111

F

(-x)(y)(-z) + (-x)yz + x(-y)z + xyz

F(x, y, z) = 1 if and only if:

7

Representing Boolean Functions

How to construct a Boolean expression that represents a Boolean Function ?

1110

1010

0100

1000

zyx

0001

0101

0011

1111

FF(x, y, z) =

x·y·z +

x·y·z +

x·y·z +

x·y·z

8

Boolean Identities

Double complement:x = x

Idempotent laws:x + x = x, x · x = x

Identity laws:x + 0 = x, x · 1 = x

Domination laws:x + 1 = 1, x · 0 = 0

Commutative laws:x + y = y + x, x · y = y ·

x

Associative laws:x + (y + z) = (x + y) + zx · (y · z) = (x · y) · z

Distributive laws:x + y ·z = (x + y)·(x + z)x · (y + z) = x ·y + x ·z

De Morgan’s laws:(x · y) = x + y, (x + y) =

x · yAbsorption laws:x + x ·y = x, x · (x + y) =

x

the Unit Property: x + x = 1 and Zero Property: x ·x = 0

9

Boolean Identities

Absorption law:

Show that x ·(x + y) = x

1) x ·(x + y) = (x + 0) ·(x + y) identity2) = x + 0 ·y distributive3) = x + y · 0 commutative4) = x + 0 domination5) = x identity

10

Dual Expression (related to identity pairs)

The dual ed of a Boolean expression e is obtained by exchanging + with , and 0 with 1 in e.

Example:e = xy + zwe = x + y + 0

Duality principle: e1e2 iff e1de2

d

x (x + y) = x iff x + xy = x (absorption)

ed =(x + y)(z + w)

ed = x · y · 1

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Dual Function

The dual of a Boolean function F represented by a Boolean expression is the function represented by the dual of this expression.

The dual function of F is denoted by Fd

The dual function, denoted by Fd, does not depend on the particular Boolean expression used to represent F.

12

Recall: Boolean Identities

Double complement:x = x

Idempotent laws:x + x = x, x · x = x

Identity laws:x + 0 = x, x · 1 = x

Domination laws:x + 1 = 1, x · 0 = 0

Commutative laws:x + y = y + x, x · y = y ·

x

Associative laws:x + (y + z) = (x + y) + zx · (y · z) = (x · y) · z

Distributive laws:x + y ·z = (x + y)·(x + z)x · (y + z) = x ·y + x ·z

De Morgan’s laws:(x · y) = x + y, (x + y) =

x · yAbsorption laws:x + x ·y = x, x · (x + y) =

x

the Unit Property: x + x = 1 and Zero Property: x ·x = 0

13

Boolean Expressions Sets Propositions

Identity Lawsx 0 = x, x 1 = x +, Complement Laws

x x = 1, x x = 0 ¬Associative Laws(x y) z = x (y z)(x y) z = x (y z)

Commutative Laws x y = y x, x y = y xDistributive Lawsx ( y z) = (x y) (x z)x ( y z) = (x y) (x z)

Propositional logic has operations , , and elements T and F such that the above properties hold for all x, y, and z.

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DNF: Disjunctive Normal Form

A literal is a Boolean variable or its complement.A minterm of Boolean variables x1,…,xn is a Boolean product of n literals y1…yn, where yi is either the literal xi or its complement xi.

Example:x y z + x y z + x y z

Disjunctive Normal Form: sum of products

minterms

We have seen how to develop a DNF expression for a function if we’re given the function’s “truth” table.

15

CNF: Conjunctive Normal Form

A literal is a Boolean variable or its complement.A maxterm of Boolean variables x1,…,xn is a Boolean sum of n literals y1…yn, where yi is either the literal xi or its complement xi.

Example:(x +y + z) (x + y + z)

(x + y +z)

Conjuctive Normal Form: product of sums

maxterms

16

Conjunctive Normal FormTo find the CNF representation of a Boolean function F1. Find the DNF representation of its complement F

2. Then complement both sides and apply DeMorgan’s laws to get F:

1110

1010

0100

1000

zyx

0001

0101

0011

1111

F

0

0

1

0

1

1

1

0

F

How do you build the CNF directly from the table ?

F = x·y·z + x·y·z + x·y·z + x·y·z

= (x+y+z) · (x+y+z) · (x+y+z) · (x+y+z)

F = (x·y·z) · (x·y·z) · (x·y·z) · (x·y·z)

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Functional CompletenessSince every Boolean function can be expressed in terms of ,+,¯, we say that the set of operators {,+,¯} is functionally complete.

{+, ¯} is functionally complete (no way!)

{, ¯} is functionally complete (note: x + y = xy)

NAND | and NOR ↓ are also functionally complete, each by itself (as a singleton set).

Recall x NAND y is true iff x or y (or both) are false and x NOR y is true iff both x and y are false.

xy = (x+y)