1. CS1022 Computer Programming & Principles
Lecture 9.1
Boolean Algebra (1)

2. Plan of lecture Introduction Boolean operations
Boolean algebra vs. logic
Boolean expressions
Laws of Boolean algebra
Boolean functions
Disjunctive normal form
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3. Introduction Boolean algebra Is a logical calculus
Structures and rules applied to logical symbols
Just like ordinary algebra is applied to numbers
Uses a two-valued set {0, 1}
Operations: conjunction, disjunction and negation
Is related with
Propositional logic and
Algebra of sets
Underpins logical circuits (electronic components)
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4. Boolean operations (1)  1  1 1 0 1
Boolean algebra uses set B  0, 1 of values
Operations:
Disjunction (),
Conjunction () and
Negation ()  1  1 1 0 1
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5. Boolean operations (2) p q p  q p  q 1 p p 1
If p and q are Boolean variables
“Variables” because they vary over set B  0, 1
That is, p, q  B
We can build the following tables: p q
p  q
p  q 1 p p 1
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6. Boolean algebra vs. logic
Boolean algebra is similar to propositional logic
Boolean variables p, q,... like propositions P, Q,...
Values B  0, 1 like F (false) and T (true)
Operators are similar:
p  q similar to P or Q
p  q similar to P and Q
p similar to not P
Tables also called “truth tables” (as in logic)
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7. Boolean expressions
As in logic, we can build more complex constructs from basic variables and operators ,  and 
Complex constructs are called Boolean expressions
A recursive definition for Boolean expressions:
All Boolean variables (p, q, r, s, etc.) are Boolean exprns.
If α and β are Boolean expressions then
α is a Boolean expression (and so is β)
(α  β) is a Boolean expression
(α  β) is a Boolean expression
Example: ((p  q)  (r  s))
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8. Equivalence of expressions
Two Boolean expressions are equivalent if they have the same truth table
The final outcome of the expressions are the same, for any value the variables have
We can prove mechanically if any two expressions are equivalent or not
“Mechanic” means “a machine (computer) can do it”
No need to be creative, clever, etc.
An algorithm explains how to prove equivalence
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9. Laws of Boolean algebra (1)
Some equivalences are very useful
They simplify expressions, making life easier
They allow us to manipulate expressions
Some equivalences are “laws of Boolean algebra”
We have, in addition to truth-tables, means to operate on expressions
Because equivalence is guaranteed we won’t need to worry about changing the meaning
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10. Laws of Boolean algebra (2)
Commutative Laws
p  q  q  p
p  q  q  p
Associativity Laws
p  (q  r)  (p  q)  r
p  (q  r)  (p  q)  r
Distributive Laws
p  (q  r)  (p  q)  (p  r)
p  (q  r)  (p  q)  (p  r)
Idempotent Laws
p  p  p
p  p  p
Absorption Laws
p  (p  q)  p
p  (p  q)  p
De Morgan’s Laws
(p  q)  p  q
(p  q)  p  q
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11. Laws of Boolean algebra (3)
Do they look familiar?
Logics, sets & Boolean algebra are all related
Summary of correspondence:
Logical operation
Set operation
Boolean operation
not   or  
and  
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12. Laws of Boolean algebra (4)
Very important:
Laws shown with Boolean variables (p, q, r, etc.)
However, each Boolean variable stand for an expression
Wherever you see p, q, r, etc., you should think of “place holders” for any complex expression
A more “correct” way to represent e.g. idempotent law:
    
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13. Laws of Boolean algebra (5)
What do we mean by “place holders”?
The laws are “templates” which can be used in many different situations
For example:
((p  q)  (q  s))  ((p  q)  (q  s))
   
((p  q)  (q  s))
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14. Laws of Boolean algebra (6)
We should be able to prove all laws
We know how to build truth tables (lectures 2.1 & 2.2)
Could you work out an algorithm to do this?
There is help at hand...
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15. Laws of Boolean algebra (7)
Using the laws of Boolean algebra, show that
(p  q)  (p  q) is equivalent to p
Solution:
(p  q)  (p  q)  (p  q)  (p  q) De Morgan’s
 (p  q)  (p  q) since p  p
 p  (q  q) Distr. Law
 p  0 since (q q)  0
 p from def. of 
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16. Boolean functions (1)
A Boolean function of n variables p1, p2, ..., pn is
A function f : Bn  B
Such that f(p1, p2, ..., pn) is a Boolean expression
Any Boolean expression can be represented in an equivalent standard format
The so-called disjunctive normal form
Format to make it easier to “process” expressions
Simpler, fewer operators, etc.
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17. Boolean functions (2) Example: function minterm m(p, q, r)
It has precisely one “1” in the final column of truth table
m(p, q, r)  1 if p  0, q  1 and r  1
We can define m(p, q, r)  p q  r
The expression p q  r is the product representation of the minterm m p q r m 1
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18. Boolean functions (3) p q r m 1
Any minterm m(p1, p2, ..., pr) can be represented as a conjunction of the variables pi or their negations
Here’s how:
Look at row where m takes value 1
If pi  1, then pi is in the product representation of m
If pi  0, then pi is in the product representation of m
We thus have m(p, q, r)  p q  r p q r m 1
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19. Disjunctive normal form (1)
We can express any Boolean function uniquely as a disjunction (a sequence of “”) of minterms
This is the disjunctive normal form
The idea is simple:
Each minterm is a conjunction (sequence of “”) which, if all holds, then we have output (result) “1”
The disjunction of minterms lists all cases which, if at least one holds, then we have output (result) “1”
In other words: we list all cases when it should be “1” and say “at least one of these is 1”
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20. Disjunctive normal form (2)
Find disjunctive normal form for (p  q)  (q r)
Let f  (p  q)  (q r). Its truth table is as below
The minterms are (rows with f  1)
p  q r
p  q r
p  q  r
p  q  r
Disjunctive normal form is
(p  q r)  (p  q r) 
(p  q  r)  (p  q  r) p q r f 1 p q r f 1 p q r f 1 p q r f 1 p q r f 1 p q r f 1
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21. Minimal set of operators
Any Boolean function can be represented with Boolean operators , , 
This is a complete set of operators
“Complete” = “we can represent anything we need”
It is not minimal, though:
De Morgan’s (p  q)  p  q, so (p  q)  (p  q)
We can define “” in terms of , 
Minimal set of Boolean operators: ,  or , 
Actually, any two operators suffice
Minimality comes at a price: expressions become very complex and convoluted
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22. Further reading
R. Haggarty. “Discrete Mathematics for Computing”. Pearson Education Ltd (Chapter 9)
Wikipedia’s entry on Boolean algebra
Wikibooks entry on Boolean algebra
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