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Show that \({\bf{x}} \oplus {\bf{y = y}} \oplus {\bf{x}}\).

Short Answer

Expert verified

The given \(x \oplus y = y \oplus x\)is proved.

Step by step solution

01

Definition

The complement of an element: \(\bar 0 = 1\) and \(\bar 1 = 0\)

The Boolean sum \( + \) or \(O{\bf{ }}R\) is \(1\) if either term is \(1\) .

The Boolean product or \(A{\bf{ }}N{\bf{ }}D\) is \(1\) if both terms are \(1\) .

The \(XOR\) operator \( \oplus \) is \(1\) if one of the terms is \(1\) (but not both).

02

Using the operator

\(x\)and \(y\) can both take on the value of \(0\) or \(1\) .

The \(XOR\) operator is \(1\) if one of the two elements (but not both) are \(1\) .

\(\begin{array}{*{20}{r}}x&{{\bf{ }}y}&{{\bf{ }}x \oplus y}&{{\bf{ }}y \oplus x}\\0&0&0&0\\0&1&1&1\\1&0&1&1\\1&1&0&0\end{array}\)

The last two columns of the table are identical.

Therefore, it get\(x \oplus y = y \oplus x\) .

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Most popular questions from this chapter

Use a \({\bf{K}}\)-map to find a minimal expansion as a Boolean sum of Boolean products of each of these functions in the variables \({\bf{w, x, y}}\) and \({\bf{z}}\).

\(\begin{array}{l}{\bf{a) wxyz + wx\bar yz + wx\bar y\bar z + w\bar xy\bar z + w\bar x\bar yz}}\\{\bf{b) wxy\bar z + wx\bar yz + w\bar xyz + \bar wx\bar yz + \bar w\bar xy\bar z + \bar w\bar x\bar yz}}\\{\bf{c) wxyz + wxy\bar z + wx\bar yz + w\bar x\bar yz + w\bar x\bar y\bar z + \bar wx\bar yz + \bar w\bar xy\bar z + \bar w\bar x\bar yz}}\\{\bf{d) wxyz + wxy\bar z + wx\bar yz + w\bar xyz + w\bar xy\bar z + \bar wxyz + \bar w\bar xyz + \bar w\bar xy\bar z + \bar w\bar x\bar yz}}\end{array}\)

Show that \({\bf{x}} \odot {\bf{y = xy + \bar x\bar y}}\).

Exercises 14-23 deal with the Boolean algebra \(\left\{ {{\bf{0,1}}} \right\}\) with addition,multiplication, and complement defined at the beginning of this section. In each case, use a table as in Example \(8\).

22. Verify the unit property.

Construct a multiplexer using AND gates, OR gates, andinverters that has as input the four bits\({{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}{\bf{,}}{{\bf{x}}_{\bf{3}}}\)and the two control bits\({{\bf{c}}_{\bf{o}}}\)and\({{\bf{c}}_{\bf{1}}}\). Set up the circuit so that\({{\bf{x}}_{\bf{i}}}\)is the output, where iis the value of the two-bit integer\({{\bf{(}}{{\bf{c}}_{\bf{1}}}{{\bf{c}}_{\bf{o}}}{\bf{)}}_{\bf{2}}}\).The depthof a combinatorial circuit can be defined by specifyingthat the depth of the initial input is 0 and if a gate has ndifferent inputs at depths\({{\bf{d}}_{\bf{1}}}{\bf{,}}{{\bf{d}}_{\bf{2}}}{\bf{,}}.....{\bf{,}}{{\bf{d}}_{\bf{n}}}\),respectively, then its outputs have depth equal to max\({\bf{(}}{{\bf{d}}_{\bf{1}}}{\bf{,}}{{\bf{d}}_{\bf{2}}}{\bf{,}}.....{\bf{,}}{{\bf{d}}_{\bf{n}}}{\bf{) + 1}}\); this value is also defined to be the depth of the gate. The depth of a combinatorial circuit is the maximum depth of the gates in the circuit.

Prove or disprove these equalities.

\(\begin{array}{l}a)\;x \oplus (y \oplus z){\bf{ = }}(x \oplus y) \oplus z\\b)\;x{\bf{ + }}(y \oplus z){\bf{ = }}(x{\bf{ + }}y) \oplus (x{\bf{ + }}z)\\c)\;x \oplus (y{\bf{ + }}z){\bf{ = }}(x \oplus y){\bf{ + }}(x \oplus z)\end{array}\)

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