Multiply out the Boolean expression (A’ + B)(A’ + C)(A’ + D).
Understand the Problem
The question is asking to multiply out the Boolean expression (A’ + B).(A’ + C).(A’ + D). This involves applying the distributive property of Boolean algebra to combine the terms.
Answer
$A' + BCD$
Answer for screen readers
The final answer is $A' + BCD$.
Steps to Solve
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Identify the Expression We start with the Boolean expression $(A' + B)(A' + C)(A' + D)$.
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Apply the Distributive Property Next, we will use the distributive property to first multiply the first two terms: $$(A' + B)(A' + C) = A' + B \cdot A' + A' \cdot C + B \cdot C$$ Since $A' \cdot A' = A'$ and $A' \cdot B = A'$, the above simplifies to: $$= A' + BC$$
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Multiply the Result with the Last Term Now, we multiply this result by the last term $(A' + D)$: $$(A' + BC)(A' + D) = A' + D \cdot A' + BC \cdot A' + BC \cdot D$$ Again, since $A' \cdot A' = A'$ and $A' \cdot BC = A'$, the result simplifies to: $$= A' + BCD$$
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Final Result Thus, the final simplified Boolean expression is: $$A' + BCD$$
The final answer is $A' + BCD$.
More Information
In Boolean algebra, simplifying expressions can help make logical operations more efficient. The distribution property is a fundamental concept that allows for combining like terms.
Tips
- Forgetting that $AA' = 0$ (the law of complementarity) which can result in incorrect simplification.
- Not recognizing that terms involving $A'$ can absorb other terms, leading to a simpler final expression than calculated.
