a. DE+DF+DG

41.

Another form of the DISTRIBUTIVE law is as shown by the lower

b. Q (R+S+T)

set of logic diagrams and truth tables in (frame 40). When input A is

c. VW+VY+VZ

at the 0 level, the truth table indicates that both input B and input C

d. JK (1 +L+M)

must be at the 1 level to obtain a 1 output from the Boolean

expression (A+B)(A+C). Therefore, (A+B)(A+C) may be expressed

as A + BC. This can be proved by applying the basic laws of

Boolean Algebra as follows:

(A + B)(A + C)

BA + AB + AC=BC

CARRYING OUT MULTIPLICATION

A + AB + AC + BC

IDEMPOTENT

A(1 + B + C) + BC

DISTRIBUTIVE

A = 1 + BC

UNION

A + BC

INTERSECTION

*When the DISTRIBUTIVE law is used, removing a variable by itself

will leave a 1, as shown below.

A + AB = (A + AB) = A(1 + B)

Convert the following Boolean expressions, using the DISTRIBUTIVE

law A + BC = (A + B)(A + C)

a. K + LM

b. (R + S)(R + T)

c. TV + X

d. J + KLM

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