To simplify $(x + y)'(x' + y')$ to a minimum number of literals, let's proceed step-by-step.

Step 1: Apply De Morgan's Theorem

Using De Morgan's theorem: $(x + y)' = x'y'$ Thus, the expression becomes: $(x'y')(x' + y')$

Step 2: Apply the Distributive Law

Now distribute $x'y'$ over $(x' + y')$: $= x'y'x' + x'y'y'$

Step 3: Apply Idempotent Law

Using the idempotent law ($A \cdot A = A$), we simplify: $= x'y' + x'y' = x'y'$

Final Answer

$(x + y)'(x' + y') = x'y'$

This expression, $x'y'$, is already in its simplest form with only two literals.

Would you like more details or have any questions?

Related Questions

1. How would this process change if the expression was $(x \cdot y)'(x' \cdot y')$?
2. Can De Morgan’s laws simplify an expression with more than two terms, such as $(x + y + z)'$?
3. How does the distributive property work in Boolean algebra compared to regular algebra?
4. How can one determine if an expression is already at its minimum literal count?
5. Why does Boolean algebra often use the idempotent and identity laws in simplification?

Tip:

In Boolean simplification, always check for laws like De Morgan’s, idempotent, and distributive early on—they can often reduce complex expressions quickly.