Discrete Mathematics CS 2610 February 17

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

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

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.

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

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

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

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

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 + 0 domination = x identity

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)

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.

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

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

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.

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

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 ?

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.