ENEE244-02xx
Digital Logic Design
Lecture 4
Announcements
• HW 1 due today.
• HW 2 up on course webpage, due on
Thursday, Sept. 18.
• “Small quiz” in recitation on Monday, Sept. 15
on material from Lectures 1,2
Agenda
• Last time:
– Error Detecting and Correcting Codes (2.11, 2.12)
– Boolean Algebra (3.1)
• This time:
– Boolean Algebra axioms and theorems (3.1, 3.2)
– Boolean Formulas and Functions (3.4)
– Canonical Formulas (3.5)
Boolean Algebra
Definition of a Boolean Algebra
• A mathematical system consisting of:
– A set of elements 𝐵 [0/1 or T/F]
– Two binary operators (+) and (⋅) [OR/AND]
– = for equivalence, () indicating order of operations
Where the following axioms/postulates hold:
– P1. Closure
For all 𝑥, 𝑦 ∈ 𝐵, 𝑥 + 𝑦 ∈ 𝐵, 𝑥 ⋅ 𝑦 ∈ 𝐵
– P2. Identity
There exist identity elements in 𝐵, denoted 0,1 relative
to (+) and (⋅), respectively.
For all 𝑥 ∈ 𝐵, 0 + 𝑥 = 𝑥 + 0 = 𝑥, 1 ⋅ 𝑥 = 𝑥 ⋅ 1 = 𝑥.
Definition of Boolean Algebra
– P3. Commutativity
The operations (+), (⋅) are commutative
For all 𝑥, 𝑦 ∈ 𝐵 𝑥 + 𝑦 = 𝑦 + 𝑥, 𝑥 ⋅ 𝑦 = 𝑦 ⋅ 𝑥
– P4. Distributivity
***Each operation (+), (⋅) is distributive over the other.
For all 𝑥, 𝑦, 𝑧 ∈ 𝐵:
𝑥 + 𝑦 ⋅ 𝑧 = 𝑥 + 𝑦 ⋅ 𝑥 + 𝑧 [𝑥 𝑂𝑅 (𝑦 𝐴𝑁𝐷 𝑧)]
𝑥 ⋅ 𝑦 + 𝑧 = 𝑥 ⋅ 𝑦 + (𝑥 ⋅ 𝑧) [𝑥 𝐴𝑁𝐷 (𝑦 𝑂𝑅 𝑧)]
Definition of Boolean Algebra
– P5. Complement
For every element 𝑥 ∈ 𝐵 there exists an element
𝑥 ∈ 𝐵called the complement of 𝑥 such that:
𝑥+𝑥 =1
𝑥⋅𝑥 =0
– P6. Non-triviality
There exist at least two elements 𝑥, 𝑦 ∈ 𝐵 such that
𝑥 ≠ 𝑦.
Principle of Duality
• (Except for P6), each postulate consists of two
expressions s.t. one expression is transformed
into the other by interchanging the operations (+)
and (⋅) as well as the identity elements 0 and 1.
• Such expressions are known as duals of each
other.
• If some equivalence is proved, then its dual is also
immediately true.
• E.g. If we prove: 𝑥 ⋅ 𝑥 + 𝑥 ⋅ 𝑥 = 1, then we
have by duality: 𝑥 + 𝑥 ⋅ 𝑥 + 𝑥 = 0
Theorems of Boolean Algebra
Theorem 3.2: Null elments:
For each element 𝑥 in a Boolean algebra:
𝑥+1=1
𝑥 ⋅ 0 = 0.
Proof:
𝑥 + 1 = 1 ⋅ 𝑥 + 1 [by P2]
= 𝑥 + 𝑥 ⋅ 𝑥 + 1 [by P5]
=𝑥+ 𝑥⋅1
[by P4]
=𝑥+𝑥
[by P2]
= 1.
by P5
Second part follows from principle of duality.
Theorems of Boolean Algebra
Theorem 3.4: The Idempotent law
For each element 𝑥 in a Boolean algebra:
𝑥+𝑥 =𝑥
𝑥⋅𝑥 =𝑥
Proof:
𝑥 + 𝑥 = 1 ⋅ 𝑥 + 𝑥 [by P2]
= 𝑥 + 𝑥 ⋅ 𝑥 + 𝑥 [by P5]
=𝑥+ 𝑥⋅𝑥
[by P4]
=𝑥+0
[by P5]
= 𝑥.
[by P2]
Second part follows from principle of duality.
Theorems of Boolean Algebra
Theorem 3.5: The Involution Law
For every 𝑥 in a Boolean algebra, 𝑥 = 𝑥.
Proof. We will use the fact that the complement is unique
(Theorem 3.1 in textbook).
𝑥 + 𝑥 = 1 [by P5]
𝑥 + 𝑥 = 1 [by P3]
𝑥 = 𝑥 by uniqueness.
Moreover,
𝑥 ⋅ 𝑥 = 0 [by P5]
𝑥 ⋅ 𝑥 = 1 [by P3]
𝑥 = 𝑥 by uniqueness.
Theorems of Boolean Algebra
Theorem 3.6: The absorption law
For each pair of elements 𝑥, 𝑦 in a Boolean algebra:
𝑥+𝑥⋅𝑦 =𝑥
𝑥⋅ 𝑥+𝑦 =𝑥
Proof:
𝑥 + 𝑥 ⋅ 𝑦 = 𝑥 ⋅ 1 + 𝑥 ⋅ 𝑦 [by P2]
= 𝑥 ⋅ 1 + 𝑦 [by P4]
= 𝑥 ⋅ 1 [by Theorem 3.2]
= 𝑥 [by P2]
Second part follows from principle of duality.
Theorems of Boolean Algebra
Theorem 3.7: DeMorgan’s Law
For each pair of elements 𝑥, 𝑦 in a Boolean algebra:
Note: Using
(𝑥 + 𝑦) = 𝑥 ⋅ 𝑦
Associativity
(𝑥 ⋅ 𝑦) = 𝑥 + 𝑦
of each
operation
Proof:
(+), (⋅).
𝑥 + 𝑦 + 𝑥 ⋅ 𝑦 = 𝑥 + 𝑦 + 𝑥 ⋅ 𝑥 + 𝑦 + 𝑦 [by P4]
Theorem 3.8
= 𝑥 + 𝑥 + 𝑦 ⋅ 𝑦 + 𝑦 + 𝑥 [by P3]
= 1 + 𝑦 ⋅ 1 + 𝑥 [by P5]
= 1 ⋅ 1 [by Theorem 3.2]
= 1 [by P2]
Analogous for multiplicative case.
Second part follows from principle of duality.
A Two-Valued Boolean Algebra
• 𝐵 = 0,1
• (+) = OR, (⋅) = AND
•
1 = 0, 0 = 1
OR
Can verify that all the postulates
hold by “perfect induction.”
[Checking that equality holds for all
possible variable settings]
AND
x
y
x
0
1
0
0
1
1
1
1
y
0
1
0
0
0
1
0
1
Boolean Formulas and Functions
• Example: 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 + 𝑦 𝑧
• Can be specified via a truth table.
X
Y
Z
f
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
0
1
0
1
0
1
1
0
0
1
1
1
1
Normal Forms
• Consider the function:
𝑓 𝑤, 𝑥, 𝑦, 𝑧 = 𝑥 + 𝑤𝑦 + 𝑤𝑦𝑧
• A literal is an occurrence of a complemented
or uncomplemented variable in a formula.
• A product term is either a literal or a product
(conjunction) of literals.
• Disjunctive normal form: A Boolean formula
written as a single product term or as a sum
(disjunction) of product terms.
Normal Forms
• Consider the function:
𝑓 𝑤, 𝑥, 𝑦, 𝑧 = 𝑧(𝑥 + 𝑦)(𝑤 + 𝑥 + 𝑦)
• A sum term is either a literal or a sum
(disjunction) of literals.
• Conjunctive normal form: A Boolean formula
written as a single sum term or as a product
(conjunction) of sum terms.
Canonical Formulas
• How to obtain a Boolean formula given a truth
table?
X
Y
Z
f
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
0
1
1
0
0
1
1
1
0
Minterm Canonical Formula
X
Y
Z
f
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
0
1
1
0
0
1
1
1
0
𝑓 𝑥, 𝑦, 𝑧 = 𝑥 𝑦 𝑧 + 𝑥𝑦 𝑧 + 𝑥 𝑦 𝑧
𝑥𝑦𝑧
𝑥𝑦𝑧
𝑥𝑦𝑧
m-Notation
X
Y
Z
f
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
0
1
1
0
0
1
1
1
0
𝑥𝑦𝑧
𝑥𝑦𝑧
𝑥𝑦𝑧
• 𝑓 𝑥, 𝑦, 𝑧 can be written as 𝑓 𝑥, 𝑦, 𝑧 = 𝑚1 +
𝑚 3 + 𝑚4
• 𝑓 𝑥, 𝑦, 𝑧 = Σ𝑚(1,3,4)
Maxterm Canonical Formula
X
Y
Z
f
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
0
1
1
0
0
1
1
1
0
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)(𝑥 + 𝑦 + 𝑧)(𝑥 + 𝑦 + 𝑧)
M-Notation
X
Y
Z
f
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
0
1
1
0
0
1
1
1
0
𝑥+𝑦+𝑧
𝑥+𝑦+𝑧
𝑥+𝑦+𝑧
𝑥+𝑦+𝑧
𝑥+𝑦+𝑧
• 𝑓 𝑥, 𝑦, 𝑧 can be written as 𝑓 𝑥, 𝑦, 𝑧 =
𝑀0 𝑀2 𝑀5 𝑀6 𝑀7
• 𝑓 𝑥, 𝑦, 𝑧 = Π𝑀(0,2,5,6,7)