Question

Let \({\bf{x}}\) and \({\bf{y}}\) belong to \(\left\{ {{\bf{0,1}}} \right\}\). Does it necessarily follow that \({\bf{x = y}}\) if there exists a value \({\bf{z}}\) in \(\left\{ {{\bf{0,1}}} \right\}\) such that,

\(\begin{array}{l}{\bf{a) xz = yz?}}\\{\bf{b) x + z = y + z?}}\\{\bf{c) x}} \oplus {\bf{z = y}} \oplus {\bf{z?}}\\{\bf{d) x}} \downarrow {\bf{z = y}} \downarrow {\bf{z?}}\\{\bf{e) x}}|{\bf{z = y}}|z{\bf{?}}\end{array}\)

A Boolean function \({\bf{F}}\) is called self-dual if and only if \({\bf{F}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{, \ldots ,}}{{\bf{x}}_{\bf{n}}}} \right){\bf{ = }}\overline {{\bf{F}}\left( {{{{\bf{\bar x}}}_{\bf{1}}}{\bf{, \ldots ,}}{{{\bf{\bar x}}}_{\bf{n}}}} \right)} \).

Short answer

a) Concluding the given equation \({\bf{x z = y z}}\).

b) Concluding the given equation \({\bf{x + z = y + z}}\).

c) It gets \({\bf{x = y}}\) for all possible values of \({\bf{z}}\) from \({\bf{x}} \oplus {\bf{ z = y}} \oplus {\bf{ z}}\).

d) Concluding the given equation \({\bf{x}} \downarrow {\bf{z = y}} \downarrow {\bf{z}}\).

e) Concluding the given equation \({\bf{x}}|{\bf{z = y}}|{\bf{z}}\).

Step-by-step solution

Step \({\bf{1}}\): Definition

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 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 \(|\) is \({\bf{1}}\) if either term is \({\bf{0}}\).

Domination laws

\(\begin{array}{l}{\bf{x + 1 = 1}}\\{\bf{x}} \bullet {\bf{0 = 0}}\end{array}\)

Check whether the given conclude that \({\bf{x = y}}\)

(a)

\({\bf{x z = y z}}\)

By the domination law: If \({\bf{z = 0}}\), then \({\bf{x z = 0}}\) and \({\bf{y z = 0}}\).

Thus, for all possible values of \({\bf{x}}\) and \({\bf{y: x z = y z}}\) holds when \({\bf{z = }}0\) (even for the case where \({\bf{x}}\) and \({\bf{y}}\) are different).

Therefore, it cannot conclude that \({\bf{x = y}}\).

Check whether the given conclude that \({\bf{x = y}}\)

(b)

\({\bf{x + z = y + z}}\)

By the domination law: If \({\bf{z = }}1\), then \({\bf{x + z = }}1\) and \({\bf{y + z = }}1\)

Thus, for all possible values of \({\bf{x}}\) and \({\bf{y: x + z = y + z}}\) holds when \({\bf{z = }}1\) (even for the case where \({\bf{x}}\) and \({\bf{y}}\) are different).

Hence, it cannot conclude that \({\bf{x = y}}\).

Check whether the given conclude that \({\bf{x = y}}\)

(c)

\({\bf{x}} \oplus {\bf{ z = y}} \oplus {\bf{ z}}\)

If \({\bf{z = }}0\), then the \({\bf{XOR}}\) operator will be \({\bf{1}}\) if the other variable is \({\bf{1}}\) and will be \({\bf{0}}\) if the other variable is \({\bf{0}}\).

\({\bf{x}} \oplus {\bf{z = y}} \oplus {\bf{z = 0}}\)

\({\bf{x}} \oplus {\bf{z = x}} \oplus 0{\bf{ = }}0\)if \({\bf{x = }}0\)

\({\bf{y}} \oplus {\bf{z = y}} \oplus 0{\bf{ = }}0\)if \({\bf{y = }}0\)

Thus \({\bf{x = y}}\).

\({\bf{x}} \oplus {\bf{z = y}} \oplus z{\bf{ = }}1\)

\({\bf{x}} \oplus {\bf{z = x}} \oplus 0{\bf{ = }}1\)if \({\bf{x = }}1\)

\({\bf{y}} \oplus {\bf{z = y}} \oplus 0{\bf{ = }}1\)if \({\bf{y = }}1\)

Thus \({\bf{x = y}}\).

If \({\bf{z = }}1\), then the \({\bf{XOR}}\) operator will be \({\bf{1}}\) if the other variable is \({\bf{0}}\) and will be \({\bf{1}}\) if the other variable is \({\bf{0}}\).

\({\bf{x}} \oplus {\bf{z = y}} \oplus {\bf{z = 0}}\)

\({\bf{x}} \oplus {\bf{z = x}} \oplus {\bf{1 = 0}}\)if \({\bf{x = 1}}\)

\({\bf{y}} \oplus {\bf{z = y}} \oplus {\bf{1 = 0}}\)if \({\bf{y = 1}}\)

Thus \({\bf{x = y}}\).

\({\bf{x}} \oplus {\bf{z = y}} \oplus {\bf{z = }}1\)

\({\bf{x}} \oplus {\bf{z = x}} \oplus {\bf{1 = }}1\)if \({\bf{x = }}0\)

\({\bf{y}} \oplus {\bf{z = y}} \oplus {\bf{1 = }}1\)if \({\bf{y = }}0\)

Thus \({\bf{x = y}}\).

So, it notes that \({\bf{x = y}}\) for all possible values of \({\bf{z}}\).

Check whether the given conclude that \({\bf{x = y}}\)

(d)

\({\bf{x}} \downarrow {\bf{z = y}} \downarrow {\bf{z}}\)

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

If \({\bf{z = }}1\), then \({\bf{x}} \downarrow {\bf{z = }}0\) and \({\bf{y}} \downarrow {\bf{z = }}0\).

Thus, for all possible values of \({\bf{x}}\) and \({\bf{y}}\) then \({\bf{x}} \downarrow {\bf{z = y}} \downarrow {\bf{z}}\) holds when \({\bf{z = }}1\) (even for the case where \({\bf{x}}\) and \({\bf{y}}\) are different).

Accordingly, it cannot conclude that \({\bf{x = y}}\).

Check whether the given conclude that \({\bf{x = y}}\)

(e)

\({\bf{x}}|{\bf{z = y}}|{\bf{z}}\)

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

If \({\bf{z = }}0\), then \({\bf{x}}|{\bf{z = 0}}\) and \({\bf{y|z = 0}}\).

Thus, for all possible values of \({\bf{x}}\) and \({\bf{y:x}} \downarrow {\bf{z = y}} \downarrow {\bf{z}}\) holds when \({\bf{z = 0}}\) (even for the case where \({\bf{x}}\) and \({\bf{y}}\) are different).

Therefore, it cannot conclude that \({\bf{x = y}}\).