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CSE 20 Lecture 9
Boolean Algebra
CK Cheng
Feb 11, 2010 Lecture notes
1
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•
P1: a+b = b+a, ab=ba
•
P2: a+bc = (a+b)(a+c)
a(b+c) = ab + ac
•
P3: a + 0 = a, a1 = a
•
P4: a + a’ = 1, a a’ = 0
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Theorem 6: for every a in B,
(a')' = a
Proof:
A is complement of a'.
The complement of a‘ is unique
Thus
a = (a')'
Theorem 7: (Absorption Law) For every pair a,b in B,
a·(a+b) = a,
a + a·b = a
Proof:
a(a+b)
a+a·b
= (a+0)(a+b)
(P3)
= a·1 + a·b
(P3)
= a+0·b
(P2)
= a(1+b)
(P2)
= a + 0
(P3)
=
a·1
(P3)
= a
(P3)
=
a
(P3)
3
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View Full Document Theorem 8
•
For every pair a, b in B
a + a’*b = a + b,a*(a’ + b) = a*b
Proof:
a + a’*b
= (a + a’)(a + b) by P2
= (1)(a + b) by P4
= (a + b) by P3
a*(a’ + b)
= a*a’ + a*b (by P3)
= 0 + a*b (by P4)
= a*b (by P3)
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Theorem 9: De Morgan’s Law
•
Theorem: For every pair a, b in set B:
(a+b)’ = a’b’, and (ab)’ = a’+b’.
•
Proof: We show that a+b and a’b’ are
complementary. According to P4, both of the
following have to be true:
(a+b) +a’b’ = 1, (a+b)(a’b’) = 0
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View Full Document Theorem 9: De Morgan’s Law (cont.)
statement
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This note was uploaded on 06/12/2011 for the course CS 1 taught by Professor Staff during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
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