Final answer:
The question involves simplifying Boolean algebra expressions by applying laws like distribution, absorption, and idempotent laws to get simplified expressions:
a) (ab + c + df)ef simplifies to abf + cef,
b) x + xy simplifies to x, and
c) (xy'+ x'z)(wx'+ yz') remains unchanged.
Step-by-step explanation:
The question is about simplifying functional expressions using Boolean algebra and its identities. Let's simplify each expression step by step:
a. (ab + c + df)ef
Using the identity for distribution: (A+B)C = AC + BC
(ab)ef + cef + dfef
Since efef=ef (using the identity idempotent law), the expression simplifies to:
abf + cef + dfef
Now, eliminate the irrelevant term dfef by using absorption law: A+AB = A
Final simplified expression is: abf + cef
b. x + xy
Using the absorption law: A+AB = A,
The simplified expression is x
c. (xy'+ x'z)(wx'+ yz')
First, distribute the terms (using distributive law), but you'll find that this expression can be left as is due to the presence of complement terms (xy' and yz' for instance, where y and y' are complements).
As such, the expression remains as: (xy'+ x'z)(wx'+ yz')
Always remain vigilant to check if the final answer makes sense and adheres to Boolean algebra laws.