Discussion
Consider the Boolean operator # with the following properties:
\(x\#0 = x\), \(x\#1 = \bar{x}\), \(x\#x = 0\) and \(x\#\bar{x} = 1\). Then \(x\#y\) is equivalent to
Answer: A
\(x\#y\) is equivalent to \(x\bar{y} + \bar{x}y\), the exclusive-OR.
The given properties are exactly those of XOR: \(x \oplus 0 = x\), \(x \oplus 1 = \bar{x}\), \(x \oplus x = 0\) and \(x \oplus \bar{x} = 1\). Check the candidate at \(y = 0\): \(x \cdot 1 + \bar{x} \cdot 0 = x\); at \(y = 1\): \(0 + \bar{x} = \bar{x}\). Both match.
\(xy + \bar{x}\bar{y}\) is XNOR, which gives \(x\#0 = \bar{x}\). \(x\bar{y} + \bar{x}\bar{y} = \bar{y}\) and \(\bar{x}y + xy = y\) ignore \(x\) entirely and fail \(x\#0 = x\).
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