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Boolean Algebra cont’ The digital abstraction

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Page 1: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Boolean Algebra cont’

The digital abstraction

��������������� ������������

����������

Page 2: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Theorem: Absorption Law

For every pair of elements a , b � B,

1. a + a · b = a

2. a · ( a + b ) = a

Proof:

(1)

abaaba ���� 1

� �ba �� 1

� �1�� ba

1�� aa�

Identity

Commutativity

Distributivity

Identity

Theorem: For any a � B, a + 1 = 1

(2) duality.

Page 3: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Theorem: Associative Law

In a Boolean algebra, each of the binary operations ( + ) and ( · ) is associative. That is, for every a , b , c � B,

1. a + ( b + c ) = ( a + b ) + c

2. a · ( b · c ) = ( a · b ) · c

Page 4: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� �� � � �� �cbacbaA ������

� �� � � �� �� �cbcbaacbaA �������Distributivity

� �� � � �� �cbaaacba �����

acabaa ���

aca��

a�

� � acbaa ���

Commutativity

Distributivity

Distributivity

Absorption Law

Absorption Law

acaba ���Idempotent Law

Proof:

(1) Let

Page 5: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� �� � � �� �� �cbcbaacbaA �������

� �� �� � � �� � � �� �ccbabcbacbcba ���������

� �� � � �� �cbabbcba �����

� � bcbab ���

bcbbba ���

bba��

bab��

b�

bcbba ���

Commutativity

Distributivity

Distributivity

Idempotent Law

Absorption Law

Commutativity

Absorption Law

Page 6: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� �� �� � � �� � � �� �ccbabcbacbcba ���������

c

Putting it all together:

� �� � � �� �� �cbcbaacbaA �������

� �� � � �� � � �� � ccbabcbaacba ���������

cba

� �cba ���

Same transitions

· before +

Page 7: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� � � �� � � �� �cbaccbabaA �������

� �� � � �� � � �� �cbaccbabcbaa ���������

� � cba ���

� � � �cbacbaA ������

(2) Duality

Also,

Page 8: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Theorem 11: DeMorgan’s Law

For every pair of elements a , b � B,

1. ( a + b )’ = a’ · b’

2. ( a · b )’ = a’ + b’

Proof:

(1) We first prove that (a+b) is the complement of a’ ·b’ .

Thus, (a+b)’ = a’ ·b’

By the definition of the complement and its uniqueness, it suffices to show: (i) (a+b)+(a’b’ ) = 1 and

(ii) (a+b)(a’b’ ) = 0.

(2) Duality (a·b)’ = a’+b’

Page 9: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� � � �� � � �� �bbaabababa ������������

� �� � � �� �bbaaab ��������

� �� � � �� �bbaaab ��������

� � � �11 ���� ab

11��

1�

Distributivity

Commutativity

Associativity

a’ and b’ are the complements of a and b respectively

Theorem: For any a � B, a + 1 = 1

Idempotent Law

Page 10: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� � � � � � � �babababa ���������

� � � �bbaaba ������

� � � �bbaaab ������

� � � �bbaaab ������

� � � �bbaaab ������

00 ������ ab

00��

0�

Commutativity

Distributivity

Commutativity

Associativity

Commutativity

a’ and b’ are the complements of a and b respectively

Theorem: For any a � B, a · 0 = 0

Idempotent Law

Page 11: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Algebra of SetsConsider a set S.

B = all the subsets of S (denoted by P(S)).

� �� ���,,SPM �

“plus” � set-union

“ times” � set-intersection

Additive identity element – empty set Ø

Multiplicative identity element – the set S.

Complement of X � B: XSX \��

Page 12: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Theorem: The algebra of sets is a Boolean algebra.

Proof:

By satisfying the axioms of Boolean algebra:

• B is a set of at least two elements

For every non empty set S:

|B| � 2.

� �SPS �,

• Closure of (

) and ( ) over B (functions ) . BBB ��

., SYX �� definitionby )(SPX �

definitionby )(SPY�

definitionby )( and SPYXSYX ����

definitionby )( and SPYXSYX ����

Page 13: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

A1. Cummutativity of (

) and ( ).

YxXxxYX ���� or :

XxYxxXY ���� or :

An element lies in the union precisely when it lies in one of the two sets X and Y. Equally an element lies in the union precisely when it lies in one of the two sets X and Y. Hence,

YX �

XY�

XYYX ���

YxXxxYX ���� and :

XxYxxXY ���� and :

XYYX ���

Page 14: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

A2. Distributivity of (

) and ( ).

� � � � � �ZXYXZYX ������

� � �.ZYXx ���

Xx� ZYx ��

Zx�Yx�

If ,Yx� Xx� Yx� YXx ��

YXx �� ZXx �� � � � �ZXYXx ����

Let

and

or

We have and . Hence,

If ,Zx� Xx� Zx� ZXx ��We have and . Hence,

or

� � � � � �ZXYXZYX ������

Page 15: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� This can be conducted in the same manner as

.

We present an alternative way:

Definition of intersection XYX �� XZX ��and

� � � � XZXYX ����

Also, definition of intersection YYX ��

definition of union ZYY ��ZYYX ���

Similarly, ZYZX ���

� � � � ZYZXYX �����

*

**

Page 16: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� � � � � �ZYXZXYX ������

Taking (* ) and (** ) we get,

�� � � � � �ZXYXZYX ������

Distributivity of union over intersection can be conducted in the same manner.

� � � � � �ZXYXZYX ������

Page 17: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

A3. Existence of additive and multiplicative identity element.

XXXSX ������ .

XXSSXSX ������ .

identity additive -

identity tivemultiplica - S

A4. Existence of the complement.

XSXBX \. ����

� � � � SXXSXSXBX ������ \\.

� � � � ������ XXSXSXBX \\.

Algebra of sets is Boolean algebra.All axioms are satisfied

Page 18: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Dual transformation - Recursive definition:Dual: expressions expressionsbase: 0 1

1 0a a , a � B\{ 0,1}

recursion step: Let E1 and E2 be Boolean expressions.Then,

E1’ [dual(E1)]’( E1 + E2 ) [ dual(E1) · dual(E2) ] ( E1 · E2 ) [ dual(E1) + dual(E2) ]

Boolean expression - Recursive definition:base: 0 , 1 , a � B – expressions.recursion step: Let E1 and E2 be Boolean expressions.

Then,E1’( E1 + E2 ) ( E1 · E2 )

Page 19: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Proof:

Let f ( x1 , x2 , …, xn ) be a Boolean expression.

We show that applying the complement on the whole expression together with replacing each variable by it’s complement, yields the dual transformation definition.

Let fd be the dual of a function f ( x1 , x2 , …, xn )

Lemma: In switching algebra, fd = f’ ( x1’ , x2’ , …, xn’ )

Induction basis: 0 , 1 – expressions.

� � 0�xf� � �

���� ������ nxxxf ,,, 21 �

� ����� ������ nxxxf ,,, 21 �� � 1�xf

� � dfxf ������ 10�

� � dfxf ������ 01�

Page 20: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Induction hypothesis: Lemma holds for Boolean expressions: E1 and E2 .

That is:

� � � � � �� �nnn xxxExxxExxxE ,,,,,,,,, 21221121 ��� ��

� � � � � �� �������������� nnn xxxExxxExxxE ,,,,,,,,, 21221121 ���

Induction step: show that it is true for E1’( E1 + E2 ) ( E1 · E2 )

� �

� � dn

dn

ExxxE

ExxxE

,2212

,1211

,,,

,,,

�����

�����

����� �� hypothesis induction � � � �� �ndnd xxxExxxE ,,,,,, 21,221,1 �� ��

���� �� LawMorganDe' � � � �� �nn xxxExxxE ���������� ,,,,,, 212211 ��

If

then,

� �nd xxxE ,,, 21 ��

Page 21: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� � � � � �� �xExExE���

21 ��

� � � � � �� �������� xExExE���

21

����� �� hypothesis induction � � � �� �xExE dd

��,2,1 ��

���� �� LawMorganDe' � � � �� �xExE ��������

21

If

then,

� �xEd

��

� � � �xExE��

1��If

� � � �xExE ��������

1then,

� �� ����� xE�

1

� �� ��� xE d

�,1

� �xEd

��

����� �� hypothesis induction

Page 22: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Definition: A function f is called self-dual if f = fd

Lemma: For any function f and any two-valued variable A, the function g = Af + A’ fd is a self-dual.

Proof: (holds for any Boolean algebra)

� � � �dfAAfdualgdual ���

� � � �dfAdualAfdual ���

� � � �� � � � � �� �dfdualAdualfdualAdual �����

� � � �fAfA d �����

� � � � ffAAfA dd �����

� � � �dd fAffAA �����

Dual definition

Distributivity

Commutativity

Page 23: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� � � �dd fAffAA �����

dd fffAfAAA ������

dd fffAfA ����� 0

dd fffAfA ����

dd fffAAf ����

Notice that the above expression has the form:

ab + a’c +bc

where “a” =A, “b”=f, “c” = fd.

Distributivity

dd fffAfAAA ������Commutativity

A’ is the complement of A

Identity

Commutativity

Page 24: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

We now prove a stronger claim:

caabbccaabBcba �������� .,,

111 bccaabbccaab �������

� �aabccaab ������ 11

abcbcacaab ������ 11

cbaabccaab ������ 11

11 cacbaababc ������

� � � �11 ����� bcacab

11 caab ���

caab ���

Identity

a’ is the complement of a

Distributivity

Commutativity

Commutativity

Distributivity

Theorem: For any a � B, a + 1 = 1

Identity

Page 25: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

dd fffAAf ����

� � � �dfAAfdualgdual ���

dfAAf ���

caabbccaab ������

For example:

cvbf ��

� �vcbfd ��

� � � �� �vcbacvbag �����

self-dual

Page 26: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Easier proof (1) for switching algebra only: (using dual properties)

dd fffAAf ����

� � � �dfAAfdualgdual ���

Switching algebra

1 and 0 �� dff

0 and 1 �� dffOR 0�dff

0���� dfAAf

dfAAf ���Identity

Page 27: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

A = 0dd ffffgdual ������ 00)(

dd ffff ����� 10

dd ffff ���� 0

ffff dd ���� 0

df�� 0

df�

dd fffg ������� �00

0’ = 1

Identity

Commutativity

Absorption Law

Theorem: For any a � B, a · 0 = 0

Identity

Easier proof (2) for switching algebra only: (case analysis)

dd fffAAf ����

� � � �dfAAfdualgdual ���

Page 28: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

A = 1

dd ffffgdual ������ 11)(

f�

fffg d ������� �11

Page 29: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� � � �1,01,0: �f

� � decreasing monotone xf

� � � ��,0in decreasingstrictly xf �

� � � �1,in increasingstrictly �xf �

Example of a transfer function for an inverter

� � � � 0.,0 ����� xfx �

� � � � 0.1, ����� xfx �

� � � ��0, interval in the concave xf

� � � �,1 interval in theconvex is �xf

� �xf

� �xf

Page 30: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

� � -1�� �f

� � -10 ��f

� � -11 ��f

� � continuous xf �

� � 1.! 11 ����� xfx �

� � 1.! 22 ����� xfx �

1

1

slope = -1

slope = -1

x

� �xf

1x � 2x

Page 31: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

1

1�

outhighV ,

inhighV ,

outlowV ,

inlowV ,

slope = -1

slope = -1

x

� �xf

outlowinlowinhighouthigh VVVV ,,,, ���

true only if:

Page 32: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

BUT, this is not always the case.

For example:

1

1� x

� �xf

outhighV ,

inhighV ,

outlowV ,

inlowV ,

slope = -1

slope = -1

outhighinhigh VV ,, �

Moreover, in this example it can be proved that no threshold values exist, which are consistent with definition 3 from lecture notes.

Page 33: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

Using the assumption:

� � 00

2010

such that

:point a exists there

xxf

xxxx

��

f (x) = x

1

1

slope < -1

x

� �xf

1x 0x 2x

0,,,,

:start with

xVVVV outlowinlowinhighouthigh ��������

0x ��0x

� �0xf

� ���0xf

y�

x�

xy ���

Page 34: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

:set �� ����� 0,, xVV inhighinhigh

�� ����� 0,, xVV inlowinlow

� � � ����� 0,, xfVfV inlowouthigh

� � � ����� 0,, xfVfV inhighoutlow

0x ��0x

� �0xf

� ���0xf

y�

x�

����� xy

� � ��� 0xf

��� 0x

� � ��� 0xf

��� 0x

f (x) = x

1

1

slope < -1

x

� �xf

1x 0x 2x

Page 35: guy/Computer_Structure03/slides/exercise2.pdf · A a b ca b c Distributivity A a b ca a b cb c a b ca aa b c aa ab ac a ac a aa b ac Commutativity Distributivity Distributivity Absorption

��� 0, xV inhigh

��� 0, xV inlow

��� 0, xV outhigh

��� 0, xV outlow

10,02min xxxx ����

true if:

10,02minset xxxx ���� �

f (x) = x

1

1

slope < -1

x

� �xf

1x 0x 2x

outhighV ,

inhighV ,

outlowV ,

inlowV ,

slope = -1

slope = -1outlowinlowinhighouthigh VVVV ,,,, ���