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bar(A.barB+barA.B) is equivalent to...

`bar(A.barB+barA.B)` is equivalent to

A

`A.bar(B)+bar(A).B`

B

`(A+barB)(bar(A)+B)`

C

`bar(A.bar(B)+bar(AB)`

D

`(A+B).(bar(A)+B)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the expression \( \overline{(A \overline{B} + \overline{A} B)} \) step by step, we will apply De Morgan's Theorems and simplify the expression. ### Step-by-Step Solution: 1. **Identify the Expression**: We start with the expression: \[ Y = \overline{(A \overline{B} + \overline{A} B)} \] 2. **Apply De Morgan's Theorem**: According to De Morgan's Theorem, the negation of a sum is the product of the negations: \[ Y = \overline{A \overline{B}} \cdot \overline{\overline{A} B} \] 3. **Negate Each Term**: Now we will negate each term inside the expression: - For \( \overline{A \overline{B}} \): \[ \overline{A \overline{B}} = \overline{A} + B \] - For \( \overline{\overline{A} B} \): \[ \overline{\overline{A} B} = A + \overline{B} \] 4. **Combine the Results**: Now substituting back into the expression: \[ Y = (\overline{A} + B)(A + \overline{B}) \] 5. **Final Expression**: The final simplified expression is: \[ Y = (\overline{A} + B)(A + \overline{B}) \] ### Conclusion: The expression \( \overline{(A \overline{B} + \overline{A} B)} \) simplifies to \( (\overline{A} + B)(A + \overline{B}) \).
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