Boolean Algebra cont’

The digital abstraction

מבנה המחשב + מבוא למחשבים ספרתיים

2#תרגול

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

1aa

Identity

Commutativity

Distributivity

Identity

Theorem: For any a B, a + 1 = 1

(2) duality.

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

cbacbaA

cbcbaacbaA Distributivity

cbaaacba

acabaa

aca

a

acbaa

Commutativity

Distributivity

Distributivity

Absorption Law

Absorption Law

acaba Idempotent Law

Proof:

(1) Let

cbcbaacbaA

ccbabcbacbcba

cbabbcba

bcbab

bcbbba

bba

bab

b

bcbba

Commutativity

Distributivity

Distributivity

Idempotent Law

Absorption Law

Commutativity

Absorption Law

ccbabcbacbcba

c

Putting it all together:

cbcbaacbaA

ccbabcbaacba

cba

cba

Same transitions

· before +

cbaccbabaA

cbaccbabcbaa

cba

cbacbaA

(2) Duality

Also,

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’

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

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

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 \

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

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

A2. Distributivity of ( ∪ ) and ( ∩ ).

.ZYXx

Xx ZYx

ZxYx

YXx ZXx ZXYXx

ZXYXZYX

Let

and

or

If ,Yx Xx Yx YXx We have and . Hence,

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

or

ZXYXZYX

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

*

**

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

ZXYXZYX

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

ZXYXZYX

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

Dual transformation - Recursive definition:Dual: expressions → expressions base: 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 )

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.

0xf nxxxf ,,, 21

nxxxf ,,, 21 1xf

dfxf 10

dfxf 01

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

That is:

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

nnn xxxExxxExxxE ,,,,,,,,, 21221121 If

then,

nd xxxE ,,, 21

xExExE

21

xExExE

21

hypothesis induction xExE dd

,2,1

LawMorganDe ' xExE 21

If

then,

xEd

xExE

1If

xExE 1

then,

xE

1

xE d

,1

xEd

hypothesis induction

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

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

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

dd fffAAf

dfAAfdualgdual

dfAAf

caabbccaab

For example:

cvbf

vcbfd

vcbacvbag

self-dual

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

dd fffAAf

dfAAfdualgdual

Switching algebra

1 and 0 dff

0 and 1 dffOR 0dff

0 dfAAf

dfAAf Identity

A = 0dd ffffgdual 00)(

dd ffff 10

dd ffff 0

ffff dd 0

df0

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

A = 1

dd ffffgdual 11)(

f

fffg d 11

1,01,0: f

decreasing monotone xf

Example of a transfer function for an inverter

xf

xf

,0in decreasingstrictly xf

0.,0 xfx

0, interval in the concave xf

1,in increasingstrictly xf

0.1, xfx

,1 interval in theconvex is xf

-1f

-10 f

-11 f

continuous xf

1.! 11 xfx

1.! 22 xfx

1

1

slope = -1

slope = -1

x

xf

1x 2x

1

1

outhighV ,

inhighV ,

outlowV ,

inlowV ,

slope = -1

slope = -1

x

xf

outlowinlowinhighouthigh VVVV ,,,,

true only if:

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.

Using the assumption:

00

2010

such that

:point a exists there

xxf

xxxx

slope < -1

f (x) = x

1

1 x

xf

1x 0x 2x

0,,,,

:start with

xVVVV outlowinlowinhighouthigh

0x 0x

0xf

0xf

y

x

xy

:set 0,, xVV inhighinhigh

0,, xVV inlowinlow

0,, xfVfV inlowouthigh

0,, xfVfV inhighoutlow

0xf

0x

0xf

0x

f (x) = x

1

1

slope < -1

x

xf

1x 0x 2x

0x 0x

0xf

0xf

y

x

xy

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 ,,,,