1. Discrete Mathematics CS 2610 September 26

2. 2 Equal Boolean Functions Two Boolean functions F and G of degree n are equal iff for all (x 1,..x n )  B n, F (x 1,..x n ) = G (x 1,..x n ) Example: F(x,y,z) = x(y+z), G(x,y,z) = xy + xz, and F=G (recall the “truth” table we put on the board yesterday) Also, note the distributive property: x(y+z) = xy + xz via the distributive law

3. 3 Boolean Functions Two Boolean expressions e 1 and e 2 that represent the exact same function F are called equivalent F(x 1,x 2,x 3 ) 1 110 1010 0100 1000 x3x3 x2x2 x1x1 0 001 0101 0011 1111 F(x 1,x 2,x 3 ) = x 1 (x 2 +x 3 )+x 1 x 2 x 3 F(x 1,x 2,x 3 ) = x 1 x 2 +x 1 x 3 +x 1 x 2 x 3

4. 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. 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 The example from yesterday.

6. 6 Representing Boolean Functions How to construct a Boolean expression that represents a Boolean Function ? 1 110 1010 0100 0000 zyx 0 001 1101 0011 1111 F (-x)(y)(-z) + (-x)yz + x(-y)z + xyz F(x, y, z) = 1 if and only if:

7. 7 Representing Boolean Functions How to construct a Boolean expression that represents a Boolean Function ? 1 110 1010 0100 1000 zyx 0 001 0101 0011 1111 F F(x, y, z) = x · y · z + x · y · z

8. 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) + 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. 9 Boolean Identities Absorption law: Show that x·(x + y) = x 1) x·(x + y) = (x + 0) ·(x + y) identity 2) = x + 0 ·ydistributive 3) = x + y· 0commutative 4) = x + 0 domination 5) = xidentity

10. 10 Dual Expression (related to identity pairs) The dual e d of a Boolean expression e is obtained by exchanging + with , and 0 with 1 in e. Example: e = xy + zw e = x + y + 0 Duality principle: e 1  e 2 iff e 1 d  e 2 d x (x + y) = x iff x + xy = x (absorption) e d =(x + y)(z + w) e d = x · y · 1

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

12. 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. 13 Boolean Expressions  Sets  Propositions Identity Laws x  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  x Distributive Laws x  ( y  z) = (x  y)  (x  z) x  ( y  z) = (x  y)  (x  z) A Boolean Algebra is a set B with operations , , and elements 0 and 1 such that the above properties hold for all x, y, and z in B.

14. 14 DNF: Disjunctive Normal Form A literal is a Boolean variable or its complement. A minterm of Boolean variables x 1,…,x n is a Boolean product of n literals y 1 …y n, where y i is either the literal x i or its complement x i. Example: 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. 15 CNF: Conjunctive Normal Form A literal is a Boolean variable or its complement. A maxterm of Boolean variables x 1,…,x n is a Boolean sum of n literals y 1 …y n, where y i is either the literal x i or its complement x i. Example: (x +y + z)  (x + y + z) Conjuctive Normal Form: product of sums maxterms

16. 16 Conjunctive Normal Form To find the CNF representation of a Boolean function F 1.Find the DNF representation of its complement F 2.Then complement both sides and apply DeMorgan’s laws to get F: 1 110 1010 0100 1000 zyx 0 001 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)

17. 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 { , ¯} 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)