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ECE 15A
Fundamentals of Logic Design
Lecture 4
Malgorzata Marek-Sadowska
Electrical and Computer Engineering Department
UCSB
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Last time :
We have proved all
fundamental laws stated for sets.
Theorem 1
: Every statement or algebraic identity deducible from P1-
P4 remains valid if the operations (+) and ( ), and the identity
elements 0 and 1 are interchanged throughout.
Theorem 2:
For every element a in Boolean algebra B: a+a =a
and aa
= a.
Theorem 3
: For each element a in Boolean algebra B,
a + 1 = 1
and
a 0 = 0.
Theorem 4
. For each pair of elements a, b in a Boolean algebra B,
a +
ab = a
and
a(a + b) = a.
Theorem 5
. In every Boolean algebra B each of the binary operations
(+) and ()is associative. That is, for every a, b and c in B:
a + (b + c) = (a + b) + c
and
a(bc) = (ab)c

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2
3
Last time: cont.
Theorem 6.
The element a’ associated with element a
in a Boolean algebra is unique.
Theorem 7.
For every a in a Boolean algebra B,
(a’)’=a.
Theorem 8
. In any Boolean algebra, 0’ = 1 and 1’ = 0.
Theorem 9.
For every a and b in Boolean algebra B,
(ab)’ = a’ + b’
and
(a+b)’ = a’b’ [De Morgan’s laws]
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Example: 2-valued Boolean algebra
Boolean algebra on B={0,1} and operators {+,
,’}
From theorem 2 [a+a=a, aa=a]:
0+0 = 0
1+1 = 1
0
0 = 0
1
1 = 1
From Theorem 3
[a+1=1, a0=0]:
0+1=1
0
0 = 0
1+1 = 1
1
0 = 0
From Theorem 8:
0’ = 1
1’ = 0
+
0
1
0
0
1
1
1
1
0
0
0
0
1
1
0
1

3
5
Today: Boolean functions
B={0,1}
B
is a set of all tuples (b1, b2,…
bk), such that
bi is either 0 or 1 for all 1<=i<=k.
B
may be viewed as the set of all k-bit strings.
Example: k = 3. B
= {(0,0,0), (0,0,1), (0,1,0),
(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)}.
B
has a size of 2
.
k
k
3
k
k
6
Boolean function
Boolean function
f: B
-> B
, k>0, m>0.
k-bit input is mapped to an m-bit output
k
m
B
B
k
m
mapping
Domain
Co-domain
An element in a domain is assigned one
element in the co-domain.
m=1: single-output function.

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