1. Chapter 9: Boolean Algebra
Discrete Math for CS

2. Disjunction and Conjunction: ∨ ∧
Our Boolean expressions will use ' to represent negation.
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3. Commutative: p ∧ q = q ∧ p and p ∨ q = q ∨ p
Reasoning Laws:
Commutative: p ∧ q = q ∧ p and p ∨ q = q ∨ p
Associative: p ∧ ( q ∧ r ) = ( p ∧ q ) ∧ r p ∨ ( q ∨ r ) = ( p ∨ q ) ∨ r
Distributive: p ∧ ( q ∨ r ) = ( p ∧ q ) ∨(p ∧ r ) p ∨ ( q ∧ r ) = ( p ∨ q ) ∧(p ∨ r )‏
Idempotent: p ∧ p = p and p ∨ p = p
Absorption: p ∧ ( p ∨ q ) = p and p ∨ ( p ∧ q ) = p
De Morgan: (p ∧ q ) ' = p' ∨ q' and (p ∨ q' ) = p' ∧ q'
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4. Prove the Distributive Law
Exercise:
Prove the Distributive Law
p ∧ ( q ∨ r ) = ( p ∧ q ) ∨(p ∧ r )‏
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5. Show that (p' ∧ q)' ∧ (p ∨ q) == p
Exercise:
Show that (p' ∧ q)' ∧ (p ∨ q) == p
(p' ∧ q)' ∧ (p ∨ q) = (p'' ∨ q') ∧ (p ∨ q)‏
= (p ∨ q') ∧ (p ∨ q)‏
= p ∨ (q' ∧ q)‏
= p ∨ 0
= p
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6. Boolean Function:
A boolean function, f, on n variables p1, p2, ..., pn is a function f: Bn --> B such that f(p1, p2, ..., pn) is a Boolean expression
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7. Minterm:
Suppose a Boolean expression of 3 variables, p, q and r, has only one 1 in its columns of a truth table. This expression is therefore equivalent to a conjunction of p. p', q, q', r and r' that is true for the same row and false elsewhere.
p q r f(p,q,r)
So f(p,q,r) is equivalent to p' ∧ q ∧ r.
Def: A minterm is a Boolean expression that has precisely one 1 in the
final column of its truth table.
The conjunction taht is equivalent to the minterm is called the product representation of the minterm.
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8. Observation:
Any minterm m(p1, p2, ..., pn) can be represented as a conjunction of the variables pi or their negations.
A Boolean expression can be thought of as a disjunction of minterms. Writing it so is putting it in Disjunctive Normal Form.
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9. Write the following in Disjunctive Normal Form
Exercise:
Write the following in Disjunctive Normal Form
f(p,q,r) = (p ∧ q) ∨ (q' ∧ r)‏
p q r f
p' ∧ q' ∧ r
p ∧ q' ∧ r
p ∧ q ∧ r'
p ∧ q ∧ r
f(p,q,r) = (p'∧q'∧r) ∨ (p∧q'∧r) ∨ (p∧q∧r') ∨ (p∧q∧r)
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10. Complete Set of Operators
A Complete Set of Operators is a set of operators in which you can express every Boolean Expression.
The Disjunctive Normal Form construction shows us that (∨,∧,' ) is a Complete Set of Operators.
Show (∧,' ) is a Complete Set of Operators.
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11. Show NAND is a Complete Set of Operators.
It suffices to show we can write p', p ∨ q and p ∧ q using only NAND.
p' = (p ∧ p)' = p NAND p
p ∨ q = (p' ∧ q')' = ((p NAND p) ∧ (q NAND q))' = ((p NAND p) NAND (q NAND q)) p ∧ q = ((p q)')' = (p NAND q) ' = (p NAND q) NAND (p NAND q)‏
Discrete Math for CS