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
IT 0344
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