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

  1. Identify the Expression We start with the Boolean expression $(A' + B)(A' + C)(A' + D)$.

  2. 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: $$= A' + BC$$

  3. 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: $$= A' + BCD$$

  4. 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.
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