Notes of Boolean Algebra, Lecture notes of Introduction to Computing

Notes on Boolean Algebra Introduction Of Computing

Typology: Lecture notes

2018/2019

Uploaded on 04/19/2019

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Boolean Algebra
A + 0 = A A + A’ = 1
A . 1 = A A. A’ = 0
1 + A = 1 A + B = B + A
0. A = 0 A . B = B . A
A + (B + C) = (A + B) + C
A. (B. C) = (A. B). C
A + A = A
A . A = A
A. (B + C) = A.B + A.C Distributive Law
A + B.C = (A+B). (A+C)
A . B = A + B De Morgan’s theorem
A + B = A . B
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Boolean Algebra

A + 0 = A A + A’ = 1

A. 1 = A A. A’ = 0

1 + A = 1 A + B = B + A

0. A = 0 A. B = B. A

A + (B + C) = (A + B) + C

A. (B. C) = (A. B). C

A + A = A

A. A = A

A. (B + C) = A.B + A.C Distributive Law A + B.C = (A+B). (A+C) A. B = A + B De Morgan’s theorem A + B = A. B

De Morgan’s theorem

A. B = A + B

A + B = A. B

Thus, is equivalent to

Verify it using truth tables. Similarly,

is equivalent to

These can be generalized to more than two

variables: to

A. B. C = A + B + C

A + B + C = A. B. C

Example 2 F = A.B.C + A.B.C + A.B.C + A.B.C = A.B.C + A.B.C + A.B.C + A.B.C + A.B.C + A.B.C = (A.B.C + A.B.C) + (A.B.C + A.B.C) + (A.B.C + A.B.C) = (A + A). B.C + (B + B). C.A + (C + C). A.B = B.C + C.A + A.B Example 3 Show that A + A.B = A A + AB = A.1 + A.B = A. (1 + B) = A. 1 = A