guy/computer_structure03/slides/exercise2.pdf · a a b ca b c distributivity a a b ca a b cb c a b...
TRANSCRIPT
Boolean Algebra cont’
The digital abstraction
��������������� ������������
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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.
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 ( ).
� � � � � �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 ������
� 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 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 )
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�
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 ��
� � � � � �� �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
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 0�dff
0���� dfAAf
dfAAf ���Identity
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 ���
�
A = 1
dd ffffgdual ������ 11)(
�
f�
fffg d ������� �11
� � � �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
� � -1�� �f
� � -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
�
��
f (x) = x
1
1
slope < -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
0x ��0x
� �0xf
� ���0xf
y�
x�
����� xy
� � ��� 0xf
��� 0x
� � ��� 0xf
��� 0x
f (x) = x
1
1
slope < -1
x
� �xf
1x 0x 2x
��� 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 ,,,, ���