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In the Boolean algebra : overline(A).ove...

In the Boolean algebra : `overline(A).overline(B) = ….`

A

`overline(A +B)`

B

`A.B`

C

`overline(overline(A)+ overline(B))`

D

`A + B`

Text Solution

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The correct Answer is:
To solve the Boolean algebra expression `overline(A) . overline(B)`, we will use De Morgan's Theorem, which states that the negation of a conjunction is the disjunction of the negations. The expression can be simplified step by step as follows: ### Step-by-Step Solution: 1. **Understanding the Expression**: We start with the expression `overline(A) . overline(B)`, which means the logical AND of the negations of A and B. 2. **Applying De Morgan's Theorem**: According to De Morgan's Theorem, the expression `overline(A) . overline(B)` can be rewritten as `overline(A + B)`. This means that the negation of the conjunction (AND operation) is equivalent to the disjunction (OR operation) of the negations. 3. **Final Expression**: Therefore, we can conclude that: \[ \overline(A) . \overline(B) = \overline(A + B) \] ### Summary: The expression `overline(A) . overline(B)` simplifies to `overline(A + B)` according to De Morgan's Theorem.
To solve the Boolean algebra expression `overline(A) . overline(B)`, we will use De Morgan's Theorem, which states that the negation of a conjunction is the disjunction of the negations. The expression can be simplified step by step as follows: ### Step-by-Step Solution: 1. **Understanding the Expression**: We start with the expression `overline(A) . overline(B)`, which means the logical AND of the negations of A and B. 2. **Applying De Morgan's Theorem**: ...
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Knowledge Check

  • What will be the input of A and B for the Boolean expression overline((A+B)).overline((A+B))=1

    A
    0,0
    B
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    C
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    D
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  • What will be the input of A and B for the Boolean expression overline((A+B)).overline((A+B))=1

    A
    0
    B
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    C
    1,0
    D
    1,1

  • The input of A and B for the Boolean expression (overline(A+B)).(overline(A.B))=1 is.

    A
    `0,0`
    B
    `0,1`
    C
    `1,0`
    D
    `1,1`

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