Discrete Mathematics CS 2610

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Discrete Mathematics CS 2610
February 17
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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(yz), G(x,y,z) xy xz,
  and FG (recall the truth table from an earlier
  slide)
• Also, note the distributive property
• x(yz) xy xz via the
  distributive law

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Boolean Functions

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

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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
• Weve expanded the initial expression into its
  sum of products form.

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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

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Representing Boolean Functions

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

F
F(x, y, z) 1 if and only if
(-x)(y)(-z) (-x)yz x(-y)z xyz
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Representing Boolean Functions

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

F
F(x, y, z) xyz xyz xyz xyz
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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 Morgans 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
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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

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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)
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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.

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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 Morgans 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
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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.

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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 y1yn, where yi is
  either the literal xi or its complement xi.
• Example

Disjunctive Normal Form sum of products
We have seen how to develop a DNF expression for
a function if were given the functions truth
table.
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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 y1yn, where yi is
  either the literal xi or its complement xi.
• Example

Conjuctive Normal Form product of sums
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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 DeMorgans
  laws to get F

How do you build the CNF directly from the table
?
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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!)

• ?, 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.