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Best answer
33 votes
33 votes

The expression will be

$f = \overline{ \overline{( x \overline{y} )}(yz)} = \overline{( \overline{x}+y)(yz)} = \overline{ \overline{x}yz+yz } = \overline{(\overline{x}+1)(yz) } = \overline{1(yz)} = \overline{yz} = \overline{y}+\overline{z}$


The final expression only contains $y$ and $z$,

Therefore, answer will be (A) f is Independent of $x$.

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12 votes
12 votes

Simplify the given circuit, as dotted input OR gate is equal to NAND gate:

(A) is the correct option!

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1 votes
1 votes

A

This can be explicitly written as $xy’ + y’+z$

which is $y’(x+1) + z$

which is $y’ + z$ since x+1 is always 1

Answer:

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