Boolean Algebra Introduction to Relations

We identified earlier Hierarchy from symbols to knowledge Levels connected by relations

Will now begin to discuss such relations Simplest relation is unary

These really don’t make much sense We have noted numbers are relevant only

When meaning assigned Context specified

Such an assignment is essential first step Consequences of actions are limited The rules of arithmetic

Specify relations between two objects We have the basic relationships of

Add subtract multiply and divide These are simple and we’re quite familiar with them

To be able to perform more complex systems of reasoning We can also talk about logical relationships

Like arithmetic We begin by talking about relations between two things

From these can go to n-ary relations

With such relations Can form a complete system for reasoning

Introduction to Algebra We will now form what we call an algebra Formally algebra or algebraic system defined as

• Set A together with one or more n-ary operations on A We will define a set with several binary relations

• Which satisfy specified set of axioms Let’s examine these

Boolean Algebra Boolean algebra put forth by George Boole 1854

An Investigation into the Laws of Thought Boolean algebra will be our algebra In Boolean algebra will work with

Two valued variables – our set Can easily extend to multiple valued logics

Binary relations – our relations AND - • OR - +

Now we need some axioms We’ll work with the following axioms or postulates Postulates

Postulates presented by Huntington 1904 Formally

Let A be a set of elements We define an algebraic system {A, •, +, 0, 1}

• and + are the operations of AND and OR 0 and 1 are distinguished elements of A

and an equivalence relation = on A such that

I. For elements a, b, and c in A the equivalence relation is 1. Reflexive

a = a for all a in A 2. Symmetric

if a = b then b = a for all a, b in A 3. Transitive

if a = b and b = c then a = c for all a, b, and c in A 4. Substitutive

if a = b then substituting a for b in any expression will result in equivalent relation

II. Let operators • and + be defined such that if a and b are elements of A Called the closure property a • b is in A a + b is in A

III. There exists elements 0 and 1 in A such that a •1 = a a + 0 = 0

IV. The operators • and + are commutative for all a and b in A a • b = b • a a + b = b + a

V. The operators • and + are distributive for all a, b, and c in A a + (b • c) = (a + b) • (a + c) a • (b + c) = a • b + a • c

VI. For every element a in A there exists an element ~a such that a • ~a = 0 a + ~a = 1

VII. There are at least 2 elements, a and b in A such that a ≠ b

Based upon these postulates Simplest algebra consists of

The set of elements A = {0, 1}

Operations • and + defined as • 0 1

0 0 0 1 0 1

+ 0 1 0 0 1 1 1 1

Discussion and justification for each given in text For our proposed algebra

We will demonstrate that each of these postulates holds

Let’s begin Using the set A

I and II are satisfied Based upon the members in set and {0,1} Definitions of • and + given above

III is satisfied because the elements of the set are defined and Definitions of the operators • and + are specified above IV is satisfied from the definitions of the operators • and +

V is demonstrated using a truth table proof Consider one example

LHS RHS

b • c a + b • c (a + b) • (a + c) a + b a + c a b c

0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1

Observe left and right hands sides the same

VI given by exhaustive proof a • ~a = 0 a + ~a = 1

Using the definitions for the operators • and +

Let a = 0 Then for the set A

~a must be 1 since 0 and 1 are the only elements in A a • ~a = 0 • ~0 = 0 • 1 = 0 from definition of • a + ~a = 0 + ~0 = 0 + 1 = 1 from the definition of +

Let a = 1

Then for the set A ~a must be 0 since 0 and 1 are the only elements in A a • ~a = 1 • ~1 = 1 • 0 = 0 from definition of • a + ~a = 1 + ~1 = 1 + 0 = 1 from the definition of +

VII proven similarly in text

Theorems

From the basic axioms Following theorems can be derived

These are essential building blocks of all our future work Th I The elements 0 and 1 are unique

Th II For every a in A a • a = a a + a = a

Th III For every a in A

a • 0 = 0 a + 1 = 1

Th IV The elements 0 and 1 are distinct and ~0 = 1 Th V For all a and b in A

a + a•b = a a • (a + b) = a

Th VI For all a in A ~a is unique Th VII For all a in A a = ~~a Th VIII For all a, b, and c in A

a ([(a + b) + c] = a(a + b) + ac = a

Th IX For all a, b, and c in A a + (b + c) = (a + b) + c a (bc) = (ab)c

Th X For all a and b in A a + ~a•b= a + b

(a + ~a) • (a + b)

a•(~a + b) = a•b Can prove with a truth table

Th XI For all a and b in A De Morgan’s Law

~(a•b)= ~a + ~b ~(a + b) = ~a•~b Note:

~(a•b) ≠ ~a•~b De Morgan’s Law extends to any number of variables For 3 variables we have

~(a•b•c)= ~a + ~b + ~c ~(a + b + c) = ~a•~b•~c

Getting Some Practice Let’s now look at working with some of these axioms and theorems When designing logic circuits several major goals

We trade these off in the process of design One of major objectives is simplicity

Why Cost Lower failure rate Easier to build Easier to test

Can use above theorems to simplify logic expressions Let’s try a few examples Examples

ab + ~ab ~a~b + ~ab + a~b ~a + ~a~b + bc~d + b~d abc + ~abc + ~bc ab + ~ac + bc ⇒ ~a~bc + ~abc + ab~c + abc ⇒ ab + ~ac a + ~ab + ~(a + b)c + ~(a + b + c) d

Getting to Work

Now that we are armed with the great collection of Postulates Theorems

Have gotten a little practice at working with them Let’s start putting them to work Boolean Algebra

Operands Operands or variables are 2 valued Binary True

1 True iff not false

False 0 False iff not true

Note Other algebras permit multiple valued logics

Operators 3 basic kinds of operators Unary

Takes single operand Called not

Not 1 is 0 0 is 1 a is ~a, a’ ~a is the complement of a

Binary Take two operands Called the AND and the OR

Written • and + We’ve already seen these earlier

Relations Using the operators and operands

Can now begin to express relations Between and among things

We can express such relations in several ways Truth table Equation

Truth Table Table of combinations

Lists all possible values of variables

Simple case Single Variable

A has two values a01

0, 1 True, false

In tabular form

More complex cases Two Variables

A, B a b 0 0 0 1 1 0 1 1

Each can be either true or false Gives 4 combinations

A, B ⇒ 0,0; 0,1; 1,0; 1,1 In tabular form

n Variables In general Number of combinations

N = 2n n is the number of variables

Using a truth table Can use a truth table to demonstrate truth of a proposition Saw one example earlier Let’s look at another LHS RHS

a b a • b ~(a•b) ~a + ~b

0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 0 0

De Morgan’s Law

Let’s look at more complex relationship

Consider serving tea from a vending machine We have 3 items

Tea ⇒ T TLM Good Tea Bad Tea 0 0 0 0 0 1 1 0 0 1 0 1 2 0 1 0 0 1 3 0 1 1 0 1 4 1 0 0 1 0 5 1 0 1 1 0 6 1 1 0 1 0 7 1 1 1 0 1

Lemon ⇒ L Milk ⇒ M

How many variables How many combinations Can now extract from the truth table

GT = 1 = T~L~M + TL~M + T~LM BT = 1 = ~T~L~M + ~T~LM + ~TL~M + ~TLM + TLM

Minterms Two expressions written in what is called

Sum of products form Each product called minterm

Also known as implicant These are also written as

GT = Σ 4, 5, 6 BT = Σ 0, 1, 2, 3, 7

or GT = m4 + m5 + m6 BT = m0 + m1 + m2 + m3 + m7 Where 4, 5, and 6 etc. are binary equivalent of

Minterm variables 4 is 100 ⇒ T~L~M

We get these by identifying all terms For which logical expression has truth value of 1

Let’s now simplify the two expressions

GT = T~L~M + TL~M + T~LM (A+A=A) = T ~M(~L + L) + T~L(M +~M) A + ~A = 1 = T~M + T~L A•1 = A = T(~M + ~ L)

BT = ~T~L~M + ~T~LM + ~TL~M + ~TLM + TLM = ~T~L(~M + M) + ~TL(~M + M) + TLM A + ~A = 1 = ~T~L + ~TL + TLM A•1 = A = ~T(~L + L) + TLM = ~T + TLM A + ~A = 1 = ~T + LM A + ~AB = A+B

Maxterms

Maxterm Mi = ~mi We get a different form of expression if we cover 0’s

Terms for which value is False

We have now written the expressions in maxterm format Also called product of sums Can also be written as

GT = Π 0, 1, 2, 3, 7 BT = Π 4, 5, 6

or GT = M0 • M1 • M2 • M3 • M7 BT = M4 • M5 • M6

These give a listing of the terms for which the Logical expression is 0

Using De Morgan’s Law GT = 0 = (T+L+M)(T+L+~M)(T+~L+M)(T+~L+~M)(~T+~L+~M) BT = 0

= (~T+L+M)(~T+~L+M)(~T+L+~M)

Implementation Let’s now look at building these functions Looking at the equations see three functions necessary

AND OR NOT

Physical hardware used to implement called gates Probably derived from concept of gating signal

Allowing signal to pass We use 3 symbols to represent logical functions Already familiar with such notions

Resistors Capacitors Transistors

These gates given by the 3 symbols A

B C

A B C

A ~A

Although drawn with 2 inputs each Can draw with any number

Practical values typically 2, 3, or 4

Examples Consider function

F = AB + CD Can rewrite as

S = AB T = CD S

T A

F = S + T Thus A

B S

CD T

First we implement the OR part Then the two AND parts A

B

CD

F Finally we bring them together

Now let’s try B

AA~BF = A~B

Let’s now return to the tea problem

LT~L

M

T

T~M

T~M+T~LObserve

Output could have been written as T(~M+~L) T~(ML)

Logic Minimization When we design something

Want the circuit to be a simple as possible Number of reasons

Cost Power Reliability Weight Size

Algebraic Means We use the theorems and postulates discussed earlier

To do this Let’s look at some examples Examples

F = Σ 1,3,7 When we write this

How many variables do we need

We have 3 minterms Can write as

F = ~A~BC + ~ABC + ABC Observe

Choice of variables is arbitrary in this case

Simplifying

F = ~AC(~B + B) + BC(~A + A) = ~AC + BC = C(~A + B)

How do we build this now

F = Π 0,2,5,7 = (A + B + C)(A + ~B + C)(~A + B + ~C)(~A + ~B + ~C) = (A + C)(~A + ~C) = ~AC + A~C F = Σ 3,4,5 = ~ABC + A~B F = Π 3,4,6,7 = (~A + C)(~B + ~C)

Karnaugh Maps

Motivation Let’s now look at the following equation

F = A~B + AB = 2 + 3 What does this reduce to How about

G = AB~C + ABC = 6 + 7 H = A~BC + ABC = 5 + 7 ABC

0 0 0 0 1 0 0 1 2 0 1 0 3 0 1 1 4 1 0 0 5 1 0 1 6 1 1 0 7 1 1 1

I = ~ABC + ABC = 3 + 7

AB 0 0 0 1 0 1 2 1 0 3 1 1

Let’s also recall a couple of truth tables Slightly re-order

Will keep the same terms What do we see between adjacent rows ABC

0 0 0 0 1 0 0 1 3 0 1 1 2 0 1 0 6 1 1 0 7 1 1 1 5 1 0 1 4 1 0 0

AB 0 0 0 1 0 1 3 1 1 2 1 0

With new ordering Recall earlier discussion of Gray code

Single bit change between rows

Observe If I combine minterms

Any two adjacent rows in reordered truth table One variable drops out

Case I 0 + 1 = ~A 1 + 3 = B 3 + 2 = A Tricky one 2 + 0 = ~b

Case II 0 + 1 = ~A~B 1 + 3 = ~A• C etc.

2 Variable Maps Let’s put this to work

A 0 1 B 0 0 1 1 2 3

Create a matrix like map Begin with 2 variables Observe

Each cell has 2 adjacent cells Those cells differ by single variable

Cells are numbered as minterms Observe that this is a Gray sequence

Let’s go back to Case I

Let minterms 0 and 1 be True 2 and 3 be False A 0 1 B

0 1 1 1 0 0 We enter this information into the map as

The simplification F = ~A~B + ~AB = ~A Same as looking at map and combining 2 true minterms

Now let’s reverse A 0 1 B

0 0 0 1 1 1

The simplification F = A~B + AB = A Same as looking at map and combining 2 true minterms

A 0 1 B 0 1 0 1 0 1

A 0 1 B 0 0 1 1 1 0

These two maps contain minterms 1 and 2

F = ~AB + A~B 0 and 3

F = ~A~B + AB Observe

No logic simplification possible In general to be able to combine terms

They must be adjacent That is differ by single variable change

3 Variable Maps

A B

0 1 C

0 0 0 1 0 1 2 3 1 1 6 7 1 0 4 5

Let’s now extend to 3 variables The map is basically the same Observe

We again use a Gray code pattern Let’s now look at Case II again

A B

0 1 C

0 0 0 0 0 1 0 0 1 1 0 0 1 0 1 1

A B

0 1 C

0 0 0 0 0 1 0 0 1 1 0 1 1 0 0 1

A three variable table

These two reduce to A~B

For F = ∑ 4,5

AC For G = ∑ 5,7

Observe

We are combining or grouping adjacent terms

A B

0 1 C

0 0 0 0 0 1 0 0 1 1 0 1 1 0 1 1

Let’s try a more complex pattern H = ∑ 4,5,7

What does this give In practice

We do not write the 0 terms A B

0 1 C

0 0 0 1 1 1 1 1 1 0 1 1

Now try

J = ∑ 2,4,5,6

Observe Several ways of combining

Now A B

0 1 C

0 0 0 1 1 1 1 1 1 0 1 1

K = ∑ 2,4,5,7 Now

Observe Top and bottom rows adjacent A

B 0 1 C

0 0 1 1 0 1 1 1 1 0 1 1

We now have 4 terms that are adjacent This function is N = Σ 0, 1, 4, 5 N = ~A~B~C + ~A~BC + A~B~C + A~BC N = ~B (~A~C + ~AC + A~C + AC) N = ~B

4 Variables Adding one more variable not too much more complex Adds more adjacencies Now where are new adjacencies

Corners 0, 2 , 8, and 10

Edges Top and bottom Left and right

A B 00 01 11 10 CD 0 0 0 1 3 2 0 1 4 5 7 6 1 1 12 13 15 14 1 0 8 9 11 10

What’s the best way to solve this one

A B 00 01 11 10 CD 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 0

Try this

A B 00 01 11 10 CD 0 0 1 1 0 1 1 1 1 1 1 1 1 0 1 1

Now this

A B 00 01 11 10 CD 0 0 1 1 0 1 1 1 1 1 1 0 1 1

Beyond 4 Variables

Several ways to solve Can continue as we are

Must be aware of new adjacencies Can build map within map

For example 3 variable map is two 2 variable maps A B 0 1 C

0 0 0 1 1 2 3

1 0 0 1 1 2 3

Manage in parts Top map is

~A (Reduced function) Bottom map is

A(Reduced function)

Going Backwards We have seen how to take function from map Let’s see how to put on then remove in simpler form 2 Variables

Let’s begin with 2 variables F = A +~AB + AB A 0 1 B

0 0 1 1 1 1 A is bottom row

AB is minterm 3 ~AB is minterm 2

M = ∏ (0, 1, 3)

Grouping and extracting

F + A + B

3 Variables Working with more variables more interesting

F = AB + A(~B + C) A B 0 1 C

0 0 0 1 1 1 1 1 1 0 1 1

From the map

F = A Now try

G = AB~C + AC + ~BC A B 0 1 C

0 0 1 0 1 1 1 1 1 1 0 1

From the map

G = AB + ~BC

4 Variables H = A~C + AB~D + ~BCD A B 00 01 11 10 C D

0 0 1 0 1 1 1 1 1 1 1 0 1 1 1

J = ~A~B~C + ~A~C~D + BCD + ~A~BC~D

A B 00 01 11 10 C D 0 0 1 1 1 0 1 1 1 1 1 1 1 0

Don’t Care Variables

As we have seen 3 binary variables give is 8 combinations

If we have a system with 3 inputs AB

FC

Those inputs potentially can take on up to 8 values For one reason or another

Some of those values may never occur Under those circumstances

We don’t care what values they take on That is we can give them values of 0 or 1

In map to help in simplification Because we don’t care what value such terms take on

Call them don’t cares

Consider function F F = Σ 0, 1, 5 We enter these as 1’s as usual A B 0 1 C

0 0 1 1 0 1 1 1 1 0 1

As the map stands we get

F = ~A~B + ~BC

Assume combinations given by minterms 4 and 6 never occur Or if they do – they are irrelevant DC = Σ 4, 6

A B 0 1 C 0 0 1 1 0 1 1 1 x 1 0 x 1

We enter these as x’s

If we assign value of

0 for don’t care minterm 6 1 for don’t care minterm 4 With don’t care terms F reduces to F = ~B

Let’s now look at a more extensive example Design a code translator that translates from BCD to Excess 3 code

Decimal Number

BCD Inputs

Excess 3 Output

w x y z f4 f3 f2 f1 0 0 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 0 2 0 0 1 0 0 1 0 1 3 0 0 1 1 0 1 1 0 4 0 1 0 0 0 1 1 1 5 0 1 0 1 1 0 0 0 6 0 1 1 0 1 0 0 1 7 0 1 1 1 1 0 1 0 8 1 0 0 0 1 0 1 1 9 1 0 0 1 1 1 0 0

Following figure expresses block diagram

In two levels of detail

Level 0

BCD to Excess 3Encoder

BCD Excess 3

BCD to Excess 3Encoder

BCD Excess 3

DI0DI1DI2DI3

DO0DO1DO2DO3

Level 1

Let’s look at the logic for one output line

Choose f3 Note

Input minterms 10..15 are don’t cares

From the map we have F3 = B~C~D + ~BD + ~BC

A B 00 01 11 10 C D 0 0 1 1 1 0 1 1 1 1 x x x x 1 0 1 x x