edited by
5,447 views
9 votes
9 votes

The Boolean theorem $\text{AB}+\overline{\text{A}}\text{C} +\text{BC = AB} + \overline{\text{A}}\text{C}$ corresponds to

  1. $\text{(A+B)} \bullet (\overline{\text{A}} +\text{C}) \bullet \text{(B+C) = (A+B)} \bullet (\overline{\text{A}} +\text{C})$
  2. $\text{AB}+\overline{\text{A}}\text{C + BC = AB + BC}$
  3. $\text{AB}+\overline{\text{A}}\text{C + BC = (A+B)} \bullet (\overline{\text{A}} +\text{C}) \bullet \text{(B+C)}$
  4. $\text{(A+B)} \bullet (\overline{\text{A}} +\text{C}) \bullet \text{(B+C) = AB} + \overline{\text{A}}\text{C}$
edited by

2 Answers

Best answer
8 votes
8 votes

$AB+\bar{A}C+BC+\bar{A}C$ can be written as

$$AB + \bar{A}C + BC$$

because $\bar{A}C$ was written twice

Now, finding the DUAL of above expression gives

$$(A+B)\cdot(\bar{A}+C)\cdot(B+C)$$ 

Applying consensus theorem we get

$$(A+B)\cdot(\bar{A}+C)$$

hence option (a) is correct

edited by
2 votes
2 votes
This Question is Solve Redundant theorem

For min-term: AB + A'C +BC = AB + A'C

AND for max-term : ( A + B ).( A' + C ).( B + C ) = ( A + B ).(A' + C )
Answer:

Related questions

5 votes
5 votes
4 answers
1
go_editor asked Jun 12, 2016
2,851 views
Which of the following is not a valid rule of $\textsf{XOR}?$$\textsf{0 XOR 0 = 0}$$\textsf{1 XOR 1 = 1}$$\textsf{1 XOR 0 = 1}$$\textsf{B XOR B = 0}$
9 votes
9 votes
4 answers
2
go_editor asked Jun 12, 2016
4,813 views
In the expression $\overline{\text{A}}(\overline{\text{A}}+\overline{\text{B}})$ by writing the first term $\text{A}$ as $\text{A + 0}$, the expression is best simplified...
10 votes
10 votes
2 answers
3
9 votes
9 votes
1 answer
4
go_editor asked Jun 12, 2016
3,875 views
The output $\text{Y}$ of the given circuit$1$$0$$\text{X}$$\text{X}'$