(A + B)(A’ * B’) = ?

Understand the Problem

The question is asking for the simplification of a Boolean algebra expression. It involves applying Boolean algebra rules to find the result of the expression (A + B)(A’ * B’). To solve it, we will apply distribution and simplify as necessary.

Answer

$0$

Steps to Solve

Apply Distribution First, we will apply the distributive property to the expression $(A + B)(A' \cdot B')$. This involves multiplying each term in the first parentheses by each term in the second parentheses.

$$(A + B)(A' \cdot B') = A \cdot (A' \cdot B') + B \cdot (A' \cdot B')$$

Simplify Each Term Now we simplify each product. We know that $A \cdot A' = 0$ (since a variable and its complement equal zero), so:

$$A \cdot (A' \cdot B') = 0 \cdot B' = 0$$

Then, we simplify the next term:

$$B \cdot (A' \cdot B') = (B \cdot A') \cdot B'$$

Combine the Terms Now we combine the results from the previous step:

$$0 + (B \cdot A') \cdot B' = (B \cdot A') \cdot B'$$

Final Simplification Since $B \cdot B' = 0$, we can simplify further:

$$(B \cdot A') \cdot B' = B \cdot 0 = 0$$

Thus the final result simplifies to 0.

The simplified expression is $0$.

More Information

In Boolean algebra, any expression that results in $0$ indicates the absence of the condition defined by the expression. This can be significant in circuit design, where a zero output may signify that certain conditions are not met.

Tips

Forgetting that $A \cdot A' = 0$ can lead to incorrect simplifications.
Not properly applying the distribution across both terms in the expression may cause missing parts of the simplified result.

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