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In the given truth table, $f(x,y)$ represent the Boolean function.

$$\begin{array}{|c|c|c|} \hline x & y & f(x,y) \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 1 \\ \hline \end{array}$$

  1. $x \leftrightarrow y$
  2. $ x \wedge y$
  3. $x \vee y$
  4. $x \rightarrow y$
in Digital Logic by Veteran (422k points)
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$(A)$ is the correct answer.

$X$ $Y$ $X\Leftrightarrow Y$
T T T
T F F
F T F
F F T

 

1 Answer

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The given function is nothing but an EXNOR which is same as BICONDITIONAL.

[$\bigodot \equiv \Leftrightarrow$]

so option A is correct.

REFER:-> https://en.wikipedia.org/wiki/Logical_biconditional

by Active (1.2k points)

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