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Q14SE

Expert-verifiedFound in: Page 844

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**Show that each of these identities holds.**

\({\bf{a)}}\)** \({\bf{x}} \odot {\bf{x = 1}}\)**

\({\bf{b)}}\)** \({\bf{x}} \odot {\bf{\bar x = 0}}\)**

\({\bf{c)}}\)** \({\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}\)**

\({\bf{a)}}\) The given equation \({\bf{x}} \odot {\bf{x = 1}}\) holds.

\({\bf{b)}}\) The given equation \({\bf{x}} \odot {\bf{\bar x = 0}}\) holds.

\({\bf{c)}}\) The given equation \({\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}\) holds.

The complement of an element: \({\bf{\bar 0 = 1}}\) and \({\bf{\bar 1 = 0}}\).

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

The Boolean product \( \bullet \) or \({\bf{AND}}\) is \({\bf{1}}\) if both terms are \({\bf{1}}\).

The \({\bf{NOR}}\) operator \( \downarrow \) is \({\bf{1}}\) if both terms are \({\bf{0}}\).

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

The \({\bf{NAND}}\) operator \(\mid \) is \({\bf{1}}\) if either term is \({\bf{0}}\).

The Boolean operator \( \odot \) is \({\bf{1}}\) if both terms have the same value.

It is given that \({\bf{x}} \odot {\bf{x = 1}}\).

Now create a table for all possible values of \({\bf{x, y}}\) and \({\bf{z}}\).

The Boolean operator \( \odot \) is \({\bf{1}}\) if both terms have the same value.

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

Then note that the last two columns of the table are identical.

Hence, it gives \({\bf{x}} \odot {\bf{x = 1}}\).

(b)

It is given that \({\bf{x}} \odot {\bf{\bar x = 0}}\).

Now create a table for all possible values of \({\bf{x, y}}\) and \({\bf{z}}\).

The Boolean operator \( \odot \) is \({\bf{1}}\) if both terms have the same value.

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

Then note that the last two columns of the table are identical.

Hence, it gives \({\bf{x}} \odot {\bf{\bar x = 0}}\).

It is given that \({\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}\).

Now create a table for all possible values of \({\bf{x, y}}\) and \({\bf{z}}\).

The Boolean operator \( \odot \) is \({\bf{1}}\) if both terms have the same value.

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

Then note that the last two columns of the table are identical.

Hence, it gives \({\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}\).

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