05 BA2 Page 1

ECEn/CS 224

Boolean Algebra – Part 2

05 BA2 Page 2

ECEn/CS 224

InversionInversion or Complement of a Function means

all 0 outputs become 1 and all 1 outputs become 0

A B F F'

0 0 0 10 1 1 01 0 1 01 1 0 1

05 BA2 Page 3

ECEn/CS 224

Inversion

DeMorgan’s Laws

( X + Y )’ = X’ Y’ . ( X Y )’ = X’ + Y’

Proof:

X Y X' Y' X + Y (X + Y)' X'Y' X Y (XY)' X' + Y'

0 0 1 1 0 1 1 0 1 10 1 1 0 1 0 0 0 1 11 0 0 1 1 0 0 0 1 11 1 0 0 1 0 0 1 0 0

05 BA2 Page 4

ECEn/CS 224

DeMorgan’s Laws

For N variables:

(X1 + X2 + X3 + …+ Xn)’ = X1’ X2’ X3’ … Xn’

( X1 X2 X3 … Xn )’ = X1’ + X2’ + X3’ + …+ Xn’

05 BA2 Page 5

ECEn/CS 224

DeMorgan’s Laws

For complex expressions, apply DeMorgan’s Laws successively

Example:

A B F F'

0 0 0 10 1 1 01 0 1 01 1 0 1

F = A’B + AB’

F’ = (A’B + AB’)’ = (A’B)’•(AB’)’ = (A+B’)•(A’+B)

= AA’+AB+A’B’+BB’= AB+A’B’

05 BA2 Page 6

ECEn/CS 224

DeMorgan’s Laws

Another Example:

[ ( a’b + 1)(cd + e’ + 0) ]’

(a’b + 1)’ + (cd + e’ + 0)’

[(a’b)’ • 1’] + (cd)’ • (e’)’ • 0’

(a + b’) • 0 + (c’ + d’) • e • 1

Note: Expression is not simplified

05 BA2 Page 7

ECEn/CS 224

DeMorgan’s Laws - One Step Rule

[ f ( X1, X2, … XN, 0, 1, +, •) ]’ =

f ( X1 ’, X2 ’, … XN ’, 1, 0, •, +)

1. Replace all variables with the inverse2. Replace + with • and • with +3. Replace 0 with 1 and 1 with 0

Be careful of hierarchy…This is biggest source of errorsBefore beginning, surround all AND terms with parens

05 BA2 Page 8

ECEn/CS 224

Canonical Forms

05 BA2 Page 9

ECEn/CS 224

Canonical Forms

• Is this equality true?– AD + A’D’ + CD ?=? AD + A’D’ + A’C

• Hard to tell– Both look minimized…– But they look different…

• For many expressions there are multiple minimal forms

• Miminimizing is not a good way to determine equality

05 BA2 Page 10

ECEn/CS 224

Canonical Forms

• A canonical form is something written in the standard way– Two expressions which are equal will have

identical canonical forms

• Is there a standard (canonical) way of representing boolean expressions?

05 BA2 Page 11

ECEn/CS 224

Truth Tables Canonical?

A B C D f

0 0 0 0 10 0 0 1 00 0 1 0 10 0 1 1 10 1 0 0 10 1 0 1 00 1 1 0 10 1 1 1 11 0 0 0 01 0 0 1 11 0 1 0 01 0 1 1 11 1 0 0 01 1 0 1 11 1 1 0 01 1 1 1 1

f = AD + A’D’ + CD f = AD + A’D’ + A’CA B C D f

0 0 0 0 10 0 0 1 00 0 1 0 10 0 1 1 10 1 0 0 10 1 0 1 00 1 1 0 10 1 1 1 11 0 0 0 01 0 0 1 11 0 1 0 01 0 1 1 11 1 0 0 01 1 0 1 11 1 1 0 01 1 1 1 1

Same truth table -two different minimal expressions.

Are truth tables canonical forms?

05 BA2 Page 12

ECEn/CS 224

Minterm Expansions

• Truth table is unique if rows in set order

• Each row can be numbered

• Call each row’s AND term a minterm

A B C D f

0 0 0 0 1 m0

0 0 0 1 0 m1

0 0 1 0 1 m2

0 0 1 1 1 m3

0 1 0 0 1 m4

0 1 0 1 0 m5

0 1 1 0 1 m6

0 1 1 1 1 m7

1 0 0 0 0 m8

1 0 0 1 1 m9

1 0 1 0 0 m10

1 0 1 1 1 m11

1 1 0 0 0 m12

1 1 0 1 1 m13

1 1 1 0 0 m14

1 1 1 1 1 m15

f = AD + A’D’ + A’C

f (A,B,C,D) = m0 + m2 + m3 + m4 + m6 + m7 + m9 + m11 + m13 + m15

f (A,B,C,D) = m(0,2,3,4,6,7,9,11,13,15)

05 BA2 Page 13

ECEn/CS 224

Minterms

• Are these minterms of 4 variables?

abcd yesab’cd’ yesab’c no, only 3 literalsa’bc’bd no, b more than once

05 BA2 Page 14

ECEn/CS 224

Minterms have names

Note: the order of the variables is significant.

A B C f

0 0 0 1 m0 = A'B'C'

0 0 1 0 m1 = A'B'C

0 1 0 1 m2 = A'BC'

0 1 1 1 m3 = A'BC

1 0 0 1 m4 = AB'C'

1 0 1 0 m5 = AB'C

1 1 0 1 m6 = ABC'

1 1 1 1 m7 = ABC

05 BA2 Page 15

ECEn/CS 224

Minterm Expansion

• A minterm expansion is unique

f (A,B,C,D) = m(0,2,3,7)

• Useful for:– Proving equality– Shorthand for representing boolean expressions

05 BA2 Page 16

ECEn/CS 224

Maxterms are POS Equivalents to Minterms

Note that maxterms are the inverse of minterms

A B C f Minterms Maxterms

0 0 0 1 m0 = A'B'C' M0 = m0' = A+B+C

0 0 1 0 m1 = A'B'C M1 = m1' = A+B+C'

0 1 0 1 m2 = A'BC' M2 = m2' = A+B'+C

0 1 1 1 m3 = A'BC M3 = m3' = A+B'+C'

1 0 0 1 m4 = AB'C' M4 = m4' = A'+B+C

1 0 1 0 m5 = AB'C M5 = m5' = A'+B+C'

1 1 0 1 m6 = ABC' M6 = m6' = A'+B'+C

1 1 1 1 m7 = ABC M7 = m7' = A'+B'+C'

05 BA2 Page 17

ECEn/CS 224

Maxterm Expansion

• Are these maxterms of 4 variables?

a + b + c + d yesa + b’ + c + d’ yesa + b’ + d’ no, only 3 literalsa’ + b + c’ + a’ + d no, a’ more than once

05 BA2 Page 18

ECEn/CS 224

Maxterm Expansion

Any function can be written as a product of maxterms. This is called a:

Standard Product of Sums (Standard POS)

Use the 0’s for f to write the POS:

f (A,B,C) =M0 M2 M3 M4

f (A,B,C) = M(0, 2, 3, 4 )

f = (A+B+C)(A+B’+C)(A+B’+C’)(A’+B+C)

A B C f

0 0 0 0 M0

0 0 1 1 M1

0 1 0 0 M2

0 1 1 0 M3

1 0 0 0 M4

1 0 1 1 M5

1 1 0 1 M6

1 1 1 1 M7

05 BA2 Page 19

ECEn/CS 224

Maxterm Expansion

• Maxterm expansions are useful– For the same things as minterm expansions

• We will use them when we want POS instead of SOP representations

05 BA2 Page 20

ECEn/CS 224

Minterm / Maxterm Relationships

f (A,B,C) = m1 + m5 + m6 + m7

f (A,B,C) = M0 M2 M3M4

For any function, if a minterm is in the Minterm Expansion,the corresponding maxterm is not in the Maxterm Expansion.

Makes it easy to convert between SOP and POS

A B C f

0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

05 BA2 Page 21

ECEn/CS 224

Minterm Example

• Is this equation true?

AD + A’D’ + CD = AD + A’D’ + A’C

1. Write each side as a truth table2. Compare which minterms each contains

Minterm expansions shorthand truth tables…

05 BA2 Page 22

ECEn/CS 224

Minterm Example• Minimize the following:

f (A,B,C) = m1 + m5 + m7

1. Write out as SOPf = A’B’C + AB’C + ABC

2. Minimizef = B’C + AC

minterm expansions shorthand expressions…

05 BA2 Page 23

ECEn/CS 224

Maxterm ExampleMinimize the following:

f (A,B,C) = M1M5M7

1. Write out as SOPf = (A+B+C’)(A’+B+C’)(A’+B’+C’)

2. Minimizef = (B+C’)(A’+C’)

maxterm expansions shorthand expressions…

05 BA2 Page 24

ECEn/CS 224

Minterm/Maxterm Example

Convert the following to POS:

F = AB + C

1. Write minterm expansionF = m1+m3+m5+m6+m7 (use truth table if it helps)

2. Convert to maxterm expansionF = M0 • M2 • M4

3. Write POS from maxterm expansionF = (A+B+C)(A+B’+C)(A’+B+C)

4. Simplify if desiredF = (B+C)(A+C)

05 BA2 Page 25

ECEn/CS 224

Algebraic Simplification

05 BA2 Page 26

ECEn/CS 224

Algebraic SimplificationObjectives

• Why Simplify?– Simpler equations less hardware to

implement, faster, cheaper, less power

By inspection: f = A’B’C + AB’C + ABC’ + ABC

Simplifying:

f = A’B’C + AB’C + ABC’ + ABC

f = B’C + AB

XY’+XY=X

4 3-input AND gates1 4-input OR gate

2 2-input AND gates1 2-input OR gate

A B C f

0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

05 BA2 Page 27

ECEn/CS 224

Hardware Cost

• Exact cost depends on technology– nMOS, CMOS, TTL, I2L, GaAs, FPGA, …

• In general:– Fewer literals

smaller/faster/lower power circuit

– Inverters don’t count• Free in some technologies• Insignificant difference in others

05 BA2 Page 28

ECEn/CS 224

Algebraic Simplification: Which Theorems To Use?

1. Essential Identities:X + 0 = X X • 1 = XX + 1 = 1 X • 0 = 0X + X = X X • X = X( X' ) ' = XX + X' = 1 X • X' = 0

X + Y = Y + X X • Y = Y • X(X + Y) + Z = X + (Y + Z) =X + Y + Z

(XY)Z = X(YZ) = XYZ

X( Y + Z ) = XY + XZ X + YZ = ( X + Y ) ( X + Z )

3. Essential:X Y + X Y' = X ( X + Y ) ( X + Y' ) = XX + XY = X X ( X + Y ) = X

( X + Y' ) Y = XY XY' + Y = X + Y4. Useful, hard to remember, easy to re-derive:

[ f ( X1, X2, … XN, 0, 1, +, • ) ]' = f ( X1', X2', … XN', 1, 0, •, + )

2. Essential Commutative, Associative, Distributive, and DeMorgan's Laws:

Suggestions:1. Focus on yellow ones2. Create duals on

right as needed3. Be familiar with 4

05 BA2 Page 29

ECEn/CS 224

4 Methods of Algebraic Simplification

1. Combine terms2. Eliminate terms3. Eliminate literals4. Add redundant terms

05 BA2 Page 30

ECEn/CS 224

Methods of Algebraic Simplification

• Combine termsUse X Y + X Y’ = X

ABC’ + A’B’C + A’BC’ =

BC’ + A’B’C

• Eliminate terms• Eliminate literals• Add redundant terms

05 BA2 Page 31

ECEn/CS 224

Methods of Algebraic Simplification

• Combine terms• Eliminate terms

A’B + B ( A’D + C)

A’B + A’BD + BC

Use X + X Y = X

A’B + BC

• Eliminate literals• Add redundant terms

05 BA2 Page 32

ECEn/CS 224

Methods of Algebraic Simplification• Combine terms• Eliminate terms• Eliminate literals

Use X + X’ Y = X + YA’B + A’B’C’D’ + ABCD’A’( B + B’C’D’) + ABCD’A’( B + C’D’) + ABCD’A’B + A’C’D’ + ABCD’A’C’D’ + B( A’ +ACD’)A’C’D’ + B( A’ +CD’)A’B + BCD’ + A’C’D’

• Add redundant terms

05 BA2 Page 33

ECEn/CS 224

Methods of Algebraic Simplification

• Combine terms• Eliminate terms• Eliminate literals• Add redundant terms

A’B’C + AB’C + ABC

A’B’C + AB’C + AB’C + ABC

B’C + AC

05 BA2 Page 34

ECEn/CS 224

Proving an Equation is True

1. Construct a truth table for both sidesi. Tedious, not very elegantii. Computers love this method

2. Convert Both Sides to minterm/maxterm expansion

3. Manipulate one side algebraically to equal the other

4. Perform the same operation on both sides

05 BA2 Page 35

ECEn/CS 224

Perform the same operation on both sides

• Valid Operations– Complement – Construct the duals

• Invalid Operations– Add a term to both sides (not reversible)– Multiply (AND) both sides by a term (not reversible)– Subtract a term from both sides (subtraction?)

05 BA2 Page 36

ECEn/CS 224

Proving an Equation Add a term to both sides

Prove y = zAdd 1 to both sides

y + 1 = z + 1Use X + 1 = 1

1 = 1Therefore y = z

But not really, you cannot add an arbitrary term to both sides.

This does not prevent you from duplicating an existing term:

Y = Z Y+Y = Z

05 BA2 Page 37

ECEn/CS 224

Proving an Equation Multiply (AND) both sides by a term

Prove y = zMultiply both sides by 0

y • 0 = z • 0Use X • 0 = 0

0 = 0Therefore y = z

But not really, you cannot multiply both sides by an arbitrary term

But, you can multiply by a redundant existing term:

y • z = y • y • z

05 BA2 Page 38

ECEn/CS 224

Proving an Equation is False

• Show one combination of values which is false (counter example)

• Use Boolean algebra manipulation– Reduce to sum of products form– Minimize– Compare the 2 sides– Be careful – some expressions have >1

minimal forms…

05 BA2 Page 39

ECEn/CS 224

Converting English to Boolean Expressions

05 BA2 Page 40

ECEn/CS 224

The air conditioner should be turned on if and only if: - the temperature is greater than 75, - the time is between 8a.m. and 5 p.m., - and it is not a holiday.

05 BA2 Page 41

ECEn/CS 224

1. Identify phrases

F = air conditioner should be turned onA = temperature is greater than 75B = time is between 8a.m. and 5 p.mC = it is not a holiday

The air conditioner should be turned on if and only if: - the temperature is greater than 75, - the time is between 8a.m. and 5 p.m., - and it is not a holiday.

05 BA2 Page 42

ECEn/CS 224

2. Identify connective words

impliedand

=

The air conditioner should be turned on if and only if: - the temperature is greater than 75, - the time is between 8a.m. and 5 p.m., - and it is not a holiday.

05 BA2 Page 43

ECEn/CS 224

3. Construct a Boolean Expression

F = air conditioner should be turned onA = temperature is greater than 75B = time is between 8a.m. and 5 p.mC = it is not a holiday

F = A • B • C’

The air conditioner should be turned on if and only if: - the temperature is greater than 75, - the time is between 8a.m. and 5 p.m., - and it is not a holiday.

05 BA2 Page 44

ECEn/CS 224

4. Draw the Network

F = A • B • C’

The air conditioner should be turned on if and only if: - the temperature is greater than 75, - the time is between 8a.m. and 5 p.m., - and it is not a holiday.

BA

CF

05 BA2 Page 45

ECEn/CS 224

Converting English to Boolean

1. Identify phrases2. Identify connective words3. Construct a Boolean Expression4. Draw the Network

05 BA2 Page 46

ECEn/CS 224

Converting English to Boolean

• Be careful: Boolean algebra is precise, English is not.

The roads will be very slippery if it snows or rains and there is oil on the road.

F = A + BC

or

F = (A + B) C

AB C

Which is it?

05 BA2 Page 47

ECEn/CS 224

AND/OR vs. OR/AND Logic Forms

05 BA2 Page 48

ECEn/CS 224

AND/OR Logic from Truth Table

1. Write the SOP by inspection from f

A

B

C

f

f = A’B’C +AB’C + ABC’ + ABC

A B C f

0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

05 BA2 Page 49

ECEn/CS 224

AND/OR Logic from Truth Table

2. Simplify the equation

f = A’B’C +AB’C + ABC’ + ABC

f = (A + A’) B’C + AB ( C + C’)

f = AB + B’C

05 BA2 Page 50

ECEn/CS 224

AND/OR Logic from Truth Table

3. Draw the logic network

f = AB + B’C

A

B

C

f

05 BA2 Page 51

ECEn/CS 224

OR/AND Logic from Truth Table

A

B

C

f

f ‘ = A’B’C’ + A’BC’ + AB’C’f = (A + B + C)(A + B’ + C)(A’ + B + C)

1. Write a POS by inspection from f

A B C f

0 0 0 00 0 1 10 1 0 00 1 1 11 0 0 01 0 1 11 1 0 11 1 1 1

05 BA2 Page 52

ECEn/CS 224

OR/AND Logic from Truth Table

3. Simplify the equation: f = (B+C)(A+C)4. Draw the logic network

AC

BC

f

05 BA2 Page 53

ECEn/CS 224

Incompletely Specified Functions

05 BA2 Page 54

ECEn/CS 224

Incompletely Specified Functions

w

x

y

z

A

B

CFN1 N2

Known: N1 does not generateoutputs A B C with valuesof 101 or 111

The X’s represent don’t care – since those input combos never occur, you don’t care what the output is.

A B C f

0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 X1 1 0 11 1 1 X

05 BA2 Page 55

ECEn/CS 224

Incompletely Specified Functions

f1 = A’B’C + AB’C’ + ABC’f1 = A’B’C + AC’

f2 = A’B’C + AB’C’ + ABC’ + ABCf2 = A’B’C + AC’(B’ + B) + AB(C’ + C)f2 = A’B’C + AC’ + AB Worse than f1

Multiple ways to choose those X’s when minimizing. Cost of minimal solution will depend on which one chosen.

A B C f f1 f2 f3 f4

0 0 0 0 0 0 0 00 0 1 1 1 1 1 10 1 0 0 0 0 0 00 1 1 0 0 0 0 01 0 0 1 1 1 1 11 0 1 X 0 0 1 11 1 0 1 1 1 1 11 1 1 X 0 1 0 1

05 BA2 Page 56

ECEn/CS 224

Incompletely Specified Functions

f1 = A’B’C + AC’

f3 = A’B’C + AB’C’ + AB’C + ABC’f3 = A’C( B + B’) + AB’( C’ + C)f3= A’C + AB’ Better than f1

f4 = A’B’C + AB’C’ + AB’C + ABC’ + ABCf4 = (A’ + A)B’C+ AB’(C’ + ’C) + AB(C’ + C)f4 = B’C+ AB’ + AB = B’C + A Best

A B C f f1 f2 f3 f4

0 0 0 0 0 0 0 00 0 1 1 1 1 1 10 1 0 0 0 0 0 00 1 1 0 0 0 0 01 0 0 1 1 1 1 11 0 1 X 0 0 1 11 1 0 1 1 1 1 11 1 1 X 0 1 0 1

05 BA2 Page 57

ECEn/CS 224

Incompletely Specified Functions

• If the function has n don’t cares– Must solve 2n truth tables– Ouch

• A better way will be shown in a later chapter…

05 BA2 Page 58

ECEn/CS 224

Don’t Cares - Minterm Representation

Use d0 … d7 to represent don’t care minterms

F = m1 + m4 + m6 + d5 + d7

F = m( 1, 4, 6) + d( 5, 7)

A B C f

0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 X1 1 0 11 1 1 X

05 BA2 Page 59

ECEn/CS 224

Don’t Cares - Maxterm Representation

Use D0 … D7 to represent don’t care maxterms

F = M0 M2 M3 D5 D7

F = M(0, 2, 3) D(5, 7)

A B C f

0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 X1 1 0 11 1 1 X