Question

(A + A') (AB+ ABC') = AB how to prove in boolean function​

Answer 1

Answer:

Step-by-step explanation: 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

Synthesis of logic circuits

Many problems of logic design can be specified using a

truth table. Give such a table, can you design the logic

circuit?

Design a logic circuit with three inputs A, B, C and one

output F such that F=1 only when a majority of the inputs is

equal to 1.

A B C F Sum of product form

0 0 0 0 F = A.B.C + A.B.C + A.B.C + A.B.C

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1 Draw a logic circuit to generate F

Simplification of Boolean functions

Using the theorems of Boolean Algebra, the algebraic

forms of functions can often be simplified, which leads to

simpler (and cheaper) implementations.

Example 1

F = A.B + A.B + B.C

= A. (B + B) + B.C How many gates do you save

= A.1 + B.C from this simplification?

= A + B.C

A A

F

B

B C

F

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