Recent questions tagged boolean-algebra

67 67 votes
7 answers 7 answers
23.8k
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The dual of a Boolean function $F(x_1,x_2,\dots,x_n,+, .,')$, written as $F^D$ is the same expression as that of $F$ with $+$ and $⋅$ swapped. $F$ is said to be self-dual...
28 28 votes
4 answers 4 answers
12.4k
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Which of the following operations is commutative but not associative?ANDORNANDEXOR
68 68 votes
8 answers 8 answers
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What happens when a bit-string is XORed with itself $n$-times as shown:$\left[B \oplus (B \oplus ( B \oplus (B \dots n \text{ times}\right]$complements when $n$ is evenco...
44 44 votes
10 answers 10 answers
16.2k
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Which one of the following expressions does NOT represent exclusive NOR of $x$ and $y$?$xy + x′ y′$$x\oplus y′$$x′\oplus y$$x′\oplus y′$
26 26 votes
6 answers 6 answers
12.8k
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Which of the following expressions is not equivalent to $\bar{x}$?$x \text{ NAND } x$$x \text{ NOR } x$$x \text{ NAND } 1$$x \text{ NOR } 1$
59 59 votes
10 answers 10 answers
17.8k
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Define the connective $*$ for the Boolean variables $X$ and $Y$ as: $$X * Y = XY + X'Y'.$$ Let $Z = X * Y$. Consider the following expressions $P$, $Q$ and $R$.$$P : X = ...
64 64 votes
7 answers 7 answers
19.7k
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Let $f(w, x, y, z) = \sum {\left(0,4,5,7,8,9,13,15\right)}$. Which of the following expressions are NOT equivalent to $f$?P: $x'y'z' + w'xy' + wy'z + xz$Q: $w'y'z' + wx'y...
39 39 votes
5 answers 5 answers
13.6k
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A Boolean function $x’y’ + xy + x’y$ is equivalent to$x' + y'$$x + y$$x + y'$$x' + y$
38 38 votes
4 answers 4 answers
12.4k
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Let $f(A,B) = A'+B$. Simplified expression for function $f(f(x+y, y), z)$ is$x' + z$$xyz$$xy' + z$None of the above
55 55 votes
8 answers 8 answers
14.0k
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The simultaneous equations on the Boolean variables $x, y, z$ and $w$,$x + y + z = 1 $$xy = 0$$xz + w = 1$$xy + \bar{z}\bar{w} = 0$have the following solution for $x, y, ...
39 39 votes
6 answers 6 answers
16.5k
16.5k views
The operation which is commutative but not associative is:ANDOREX-ORNAND
44 44 votes
4 answers 4 answers
18.1k
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If $P, Q, R$ are Boolean variables, then$(P + \bar{Q}) (P.\bar{Q} + P.R) (\bar{P}.\bar{R} + \bar{Q})$ simplifies to$P.\bar{Q}$$P.\bar{R}$$P.\bar{Q} + R$$P.\bar{R} + Q$
47 47 votes
3 answers 3 answers
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The truth table ${\begin{array}{|c|c|c|}\hline\textbf{X}& \textbf{Y}& \textbf{(X,Y)} \\\hline0& 0& 0 \\ \hline 0& 1&0\\ \hline1& 0& 1 \\\hline1&...