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Question
Applying Dr. Morgans theorem to the expression ((x+y)'+z)', we get ______.
Options
(x + y)z
(x' + y')z
(x + y)z'
(x' + y')z'
MCQ
Fill in the Blanks
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Solution
Applying Dr. Morgans theorem to the expression ((x+y)'+z)', we get (x + y)z'.
Explanation:
Let P = (x+y)'.
Then [(P+z)'=P'z'] would appear if the expression were [((x+y)'+z)').
Since [P'=((x+y)')'=x+y],
the result is [(x+y)z'].
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