1. 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) 1 x3 x2 x1
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 ? 1 z y x F
F(x, y, z) = 1 if and only if:
(-x)(y)(-z) + (-x)yz + x(-y)z + xyz

7. Representing Boolean Functions
How to construct a Boolean expression that represents a Boolean Function ? 1 z y x F
F(x, y, z) =
x·y·z +
x·y·z

8. the Unit Property: x + x = 1 and Zero Property: x ·x = 0
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) + z
x · (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 · y
Absorption 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
x ·(x + y) = (x + 0) ·(x + y) identity
= x + 0 ·y distributive
= x + y · 0 commutative
= x domination
= 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 + zw
e = x + y + 0
ed =(x + y)(z + w)
ed = x · y · 1
Duality principle: e1e2 iff e1de2d
x (x + y) = x iff x + xy = x (absorption)

11. 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) + z
x · (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 · y
Absorption 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 Laws
x  0 = x, x  1 = x +,
Complement Laws
x  x = 1, x  x = ¬
Associative Laws
(x  y)  z = x  (y  z)
(x  y)  z = x  (y  z)
Commutative Laws
x  y = y  x, x  y = y  x
Distributive Laws
x  ( 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.

14. 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:
minterms
x y z
+ x y z
Disjunctive Normal Form: sum of products
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:
maxterms
(x +y + z)
 (x + y + z)
 (x + y +z)
Conjuctive Normal Form: product of sums

16. Conjunctive Normal Form
To find the CNF representation of a Boolean function F
Find the DNF representation of its complement F
Then complement both sides and apply DeMorgan’s laws to get F: 1 z y x F
F = 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)
= (x+y+z) · (x+y+z) · (x+y+z) · (x+y+z)
How do you build the CNF directly from the table ?

17. Functional Completeness
Since every Boolean function can be expressed in terms of ,+,¯, we say that the set of operators {,+,¯} is functionally complete.
{+, ¯} is functionally complete (no way!)
xy = (x+y)
{, ¯} 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.