Question

Exercises: Simplify to a minimum of literals.

a) BD +B (D+E) + D' (D+F)
b) A'B'C (A+B+C')' + A'B'C'D
c) (B+BC) (B+B 'C) (B+D)

Answer 1

To simplify the given Boolean expressions, apply the laws of Boolean algebra, such as the distributive law and De Morgan's law. The simplified expressions are BD + BE + DF, A'B'C + A'B'C'D, B*D.

Step-by-step explanation:

To simplify the given expressions, we can apply the laws of Boolean algebra. Let's simplify each expression one by one:

a) BD + B(D+E) + D'(D+F)

Using the distributive law, we can expand the second term: B(D+E) = BD + BE

Next, let's distribute D' in the third term: D'(D+F) = DD' + DF = 0 + DF = DF

Combining the simplified terms, the expression becomes: BD + BD + BE + DF = BD + BE + DF

b) A'B'C(A+B+C')' + A'B'C'D

Using De Morgan's law, we can simplify the first term: (A+B+C')' = A'B'C

Substituting this, the expression becomes: A'B'C + A'B'C'D

c) (B+BC)(B+B'C)(B+D)

Using the distributive law, we can simplify each term: B+BC = B, B+B'C = B, B+D = B+D

Therefore, the overall expression becomes: BBD = B*D.