Question

A Boolean function that can be represented by a threshold gate is called a threshold function. Show that each of these functions is a threshold function.

\(\begin{array}{l}a){\bf{F(x) = \bar x}}\\{\bf{b)F(x,y) = x + y}}\\{\bf{c)F(x,y) = xy}}\\{\bf{d)F(x,y) = x}}\mid y\\e)F({\bf{x,y) = }}x \downarrow y\\f)F(x,y,{\bf{z) = x + yz}}\\{\bf{g)F(w,x,y,z) = w + xy + z}}\\{\bf{h)F(w,x,y,z) = wxz + x\bar y}}z\end{array}\)

Short answer

\((a)\)The given function \({\bf{F(x) = \bar x}}\) is threshold function.

\(({\bf{b)}}\)The given function \({\bf{F(x,y) = x + y}}\) is threshold function.

\(({\bf{c)}}\)The given function \({\bf{F(x,y) = xy}}\) is threshold function.

\(({\bf{d)}}\)The given function \({\bf{F(x,y) = x}}\mid y\) is threshold function.

\((e)\)The given function \(F({\bf{x,y) = }}x \downarrow y\) is threshold function.

\((f)\)The given function \(F(x,y,{\bf{z) = x + yz}}\) is threshold function.

\(({\bf{g)}}\)The given function \({\bf{F(w,x,y,z) = w + xy + z}}\) is threshold function.

\(({\bf{h)}}\) The given function \({\bf{F(w,x,y,z) = wxz + x\bar y}}z\) is threshold function.

Step-by-step solution

Definition

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

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

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

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

The \(XOR\) 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 \({\bf{0}}\).

A threshold gate produces output \({\bf{y}}\) if the input \({{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}{\bf{, \ldots ,}}{{\bf{x}}_{\bf{n}}}\) exceeds the threshold value \(T\) for the weights \({{\bf{w}}_{\bf{1}}}{\bf{,}}{{\bf{w}}_{\bf{2}}}{\bf{, \ldots ,}}{{\bf{w}}_{\bf{n}}}\left( {{{\bf{w}}_{\bf{1}}}{{\bf{x}}_{\bf{1}}}{\bf{ + }}{{\bf{w}}_{\bf{2}}}{{\bf{x}}_{\bf{2}}}{\bf{ + \ldots + }}{w_n}{x_n} \ge T} \right)\).

Check whether the given is threshold or not

(a)

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

Let us first determine a table of values for all possible values of \({\bf{x}}\).

\(\begin{array}{*{20}{c}}x&{F(x){\bf{ = }}\bar x}\\0&1\\1&0\end{array}\)

Let us assign the following values:

\(w = - 1\)(Weight of \(x\))

\(T = - \frac{1}{2}\)

If \(x{\bf{ = 1}}\), then \(wx{\bf{ = - }}1 \cdot 1{\bf{ = - 1 < }}T\) and thus the threshold gate then returns \({\bf{0}}\).

If \({\bf{x = }}0\), then \(wx{\bf{ = - }}1 \cdot 0{\bf{ = 0 > }}T\) and thus the threshold gate then returns \(1\).

Thus, the Boolean function \({\bf{F(x) = \bar x}}\) can be represented by this threshold gate.

Hence, \({\bf{F(x) = \bar x}}\) is a threshold function.

Check whether the given is threshold or not

(b)

It is given that \({\bf{F(x,y) = x + y}}\).

Let us first determine a table of values for all possible values of \(x\).

\(\begin{array}{*{20}{c}}x&y&{F(x{\bf{,y) = x + }}y}\\0&0&0\\0&1&1\\1&0&1\\1&1&1\end{array}\)

Let us assign the following values:

\({w_x}{\bf{ = 1}}\)(Weight of \(x\))

\({{\bf{w}}_{\bf{y}}}{\bf{ = 1 }}\)(Weight of \({\bf{y}}\))

\({\bf{T = }}\frac{{\bf{1}}}{{\bf{2}}}\)

If \({\bf{x = 1}}\) and \({\bf{y = 1}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{ y = x + y = 1 + 1 = 2 > T}}\) and thus the threshold gate then returns \({\bf{1}}\).

If \({\bf{x = 1}}\) and \({\bf{y = 0}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{y = x + y = 1 + 0 = 1 > T}}\) and thus the threshold gate then returns \({\bf{1}}\).

If \(x{\bf{ = 0}}\) and \({\bf{y = 1}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{y = x + y = 0 + 1 = 1 > T}}\) and thus the threshold gate then returns \({\bf{1}}\).

If \(x{\bf{ = 0}}\) and \({\bf{y = 0}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{y = x + y = 0 + 0 = 0 < T}}\) and thus the threshold gate then returns \({\bf{0}}\).

Thus, the Boolean function \({\bf{F(x,y) = x + y}}\) can be represented by this threshold gate.

Hence, \({\bf{F(x,y) = x + y}}\) is a threshold function.

Check whether the given is threshold or not

(c)

It is given that \({\bf{F(x,y) = xy}}\).

Let us first determine a table of values for all possible values of \({\bf{x}}\).

\(\begin{array}{*{20}{c}}x&y&{F(x,y){\bf{ = }}xy}\\0&0&0\\0&1&0\\1&0&0\\1&1&1\end{array}\)

Let us assign the following values:

\({{\bf{w}}_{\bf{x}}}{\bf{ = 1}}\)(Weight of \({\bf{x}}\))

\({{\bf{w}}_{\bf{y}}}{\bf{ = }}1\)(Weight of \({\bf{y}}\))

\(T = \frac{3}{2}\)

If \({\bf{x = 1}}\) and \({\bf{y = 1}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{ y = x + y = 1 + 1 = 2 > T}}\) and thus the threshold gate then returns \({\bf{1}}\).

If \({\bf{x = 1}}\) and \({\bf{y = 0}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{y = x + y = 1 + 0 = 1 < T}}\) and thus the threshold gate then returns \({\bf{0}}\).

If \(x{\bf{ = 0}}\) and \({\bf{y = 1}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{y = x + y = 0 + 1 = 1 < T}}\) and thus the threshold gate then returns \({\bf{0}}\).

If \(x{\bf{ = 0}}\) and \({\bf{y = 0}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{y = x + y = 0 + 0 = 0 < T}}\) and thus the threshold gate then returns \({\bf{0}}\).

Thus, the Boolean function \({\bf{F(x,y) = xy}}\) can be represented by this threshold gate.

Hence, \({\bf{F(x,y) = xy}}\) is a threshold function.

Check whether the given is threshold or not

(d)

It is given that \({\bf{F(x,y) = x}}\mid y\).

Let us first determine a table of values for all possible values of \({\bf{x}}\).

\(\begin{array}{*{20}{c}}x&y&{F(x,y){\bf{ = }}x\mid y}\\0&0&1\\0&1&1\\1&0&1\\1&1&0\end{array}\)

Let us assign the following values:

\({{\bf{w}}_{\bf{x}}}{\bf{ = - 1}}\)(Weight of \({\bf{x}}\))

\({{\bf{w}}_{\bf{y}}}{\bf{ = - }}1\)(Weight of \({\bf{y}}\))

\(T = {\bf{ - }}\frac{3}{2}\)

If \({\bf{x = 1}}\) and \({\bf{y = 1}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{ y = }} - x - y = - 1 - 1 = - 2{\bf{ < T}}\) and thus the threshold gate then returns \({\bf{0}}\).

If \({\bf{x = 1}}\) and \({\bf{y = 0}}\), then \({{\bf{w}}_x}x + {w_y}y = - x - y = - 1 - 0 = - 1 > T\) and thus the threshold gate then returns \({\bf{1}}\).

If \(x{\bf{ = 0}}\) and \({\bf{y = 1}}\), then \({w_x}x + {w_y}y = - x - y = - 0 - 1 = - 1 > T\) and thus the threshold gate then returns \({\bf{1}}\).

If \(x{\bf{ = 0}}\) and \({\bf{y = 0}}\), then \({w_x}x + {w_y}y = - x - y = - 0 - 0 = 0 > {\bf{T}}\) and thus the threshold gate then returns \({\bf{1}}\).

Thus, the Boolean function \({\bf{F(x,y) = x}}\mid y\) can be represented by this threshold gate.

Hence, \({\bf{F(x,y) = x}}\mid y\) is a threshold function.

Check whether the given is threshold or not

(e)

It is given that \({\bf{F(x,y) = x}} \downarrow {\bf{y}}\).

Let us first determine a table of values for all possible values of \(x\).

\(\begin{array}{*{20}{c}}x&y&{F(x,y){\bf{ = }}x \downarrow y}\\0&0&1\\0&1&0\\1&0&0\\1&1&0\end{array}\)

Let us assign the following values:

\({{\bf{w}}_{\bf{x}}}{\bf{ = - 1}}\)(Weight of \(x\))

\({{\bf{w}}_{\bf{y}}}{\bf{ = - }}1\)(Weight of \(y\))

\(T = {\bf{ - }}\frac{1}{2}\)

If \({\bf{x = 1}}\) and \({\bf{y = 1}}\), then \({{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{ y = }} - x - y = - 1 - 1 = - 2{\bf{ < T}}\) and thus the threshold gate then returns \(0\).

If \({\bf{x = 1}}\) and \({\bf{y = 0}}\), then \({{\bf{w}}_x}x + {w_y}y = - x - y = - 1 - 0 = - 1{\bf{ < }}T\) and thus the threshold gate then returns \(0\).

If \(x{\bf{ = 0}}\) and \({\bf{y = 1}}\), then \({w_x}x + {w_y}y = - x - y = - 0 - 1 = - 1{\bf{ < }}T\) and thus the threshold gate then returns \(0\).

If \(x{\bf{ = 0}}\) and \({\bf{y = 0}}\), then \({w_x}x + {w_y}y = - x - y = - 0 - 0 = 0 > {\bf{T}}\) and thus the threshold gate then returns \({\bf{1}}\).

Thus, the Boolean function \(F({\bf{x,y) = }}x \downarrow y\) can be represented by this threshold gate.

Hence, \(F({\bf{x,y) = }}x \downarrow y\) is a threshold function.

Check whether the given is threshold or not

(f)

It is given that \(F(x,y,{\bf{z) = x + yz}}\).

Let us first determine a table of values for all possible values of \(x\).

\(\begin{array}{*{20}{c}}x&y&z&{yz}&{F(x,y,z){\bf{ = x + }}yz}\\0&0&0&0&0\\0&0&1&0&0\\0&1&0&0&0\\0&1&1&1&1\\1&0&0&0&1\\1&0&1&0&1\\1&1&0&0&1\\1&1&1&1&1\end{array}\)

Let us assign the following values:

\({{\bf{w}}_{\bf{x}}}{\bf{ = }}2\)(Weight of \(x\))

\({{\bf{w}}_{\bf{y}}}{\bf{ = }}1\)(Weight of \(y\))

\({w_z}{\bf{ = 1}}\)(Weight of \({\bf{z}}\))

\(T = \frac{3}{2}\)

Let us then finally check that the threshold is reached for those combinations that need to return \({\bf{1}}\).

\(\begin{array}{*{20}{c}}{\bf{x}}&{\bf{y}}&{\bf{z}}&{{{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{y + }}{{\bf{w}}_{\bf{z}}}{\bf{z}}}&{{\bf{Threshold}}}&{{\bf{Returnedvalues}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{{\bf{ 0 + 0 + 0 = 0}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{{\bf{ 0 + 0 + 1 = 1}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{{\bf{ 0 + 1 + 0 = 1}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{{\bf{ 0 + 1 + 1 = 2}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{{\bf{ 2 + 0 + 0 = 2}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}&{{\bf{ 2 + 0 + 1 = 3}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{{\bf{ 2 + 1 + 0 = 3}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{{\bf{ 2 + 1 + 1 = 4}}}&{{\bf{ > T}}}&{\bf{1}}\end{array}\)

Thus, the Boolean function \(F(x,y,{\bf{z) = x + yz}}\) can be represented by this threshold gate. Hence, \(F(x,y,{\bf{z) = x + yz}}\) is a threshold function.

Check whether the given is threshold or not

(g)

It is given that \({\bf{F(w,x,y,z) = w + xy + z}}\).

Let us first determine a table of values for all possible values of \(x\).

\(\begin{array}{*{20}{c}}w&x&y&z&{F(w,x,y,z) = w + xy + z}\\0&0&0&0&0\\0&0&0&1&1\\0&0&1&0&0\\0&0&1&1&1\\0&1&0&0&0\\0&1&0&1&1\\0&1&1&0&1\\0&1&1&1&1\\1&0&0&0&1\\1&0&0&1&1\\1&0&1&0&1\\1&0&1&1&1\\1&1&0&0&1\\1&1&0&1&1\\1&1&1&0&1\\1&1&1&1&1\end{array}\)

Let us assign the following values:

\({w_w} = 2\)(Weight of \({\bf{w}}\))

\({{\bf{w}}_{\bf{x}}}{\bf{ = }}1\)(Weight of \(x\))

\({{\bf{w}}_{\bf{y}}}{\bf{ = }}1\)(Weight of \(y\))

\({w_z}{\bf{ = }}2\)(Weight of \({\bf{z}}\))

\(T = \frac{3}{2}\)

Let us then finally check that the threshold is reached for those combinations that need to return \({\bf{1}}\).

\(\begin{array}{*{20}{c}}{\bf{w}}&{\bf{x}}&{\bf{y}}&{\bf{z}}&{{{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{y + }}{{\bf{w}}_{\bf{z}}}{\bf{z}}}&{{\bf{Threshold}}}&{{\bf{Returnedvalues}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{{\bf{0 + 0 + 0 + 0 = 0}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{{\bf{0 + 0 + 0 + 2 = 2}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{{\bf{0 + 0 + 1 + 0 = 1}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{{\bf{0 + 0 + 1 + 2 = 3}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{{\bf{0 + 1 + 0 + 0 = 1}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{1}}&{{\bf{0 + 1 + 0 + 2 = 3}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{{\bf{0 + 1 + 1 + 0 = 2}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{{\bf{0 + 1 + 1 + 2 = 4}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{{\bf{2 + 0 + 0 + 0 = 2}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{{\bf{2 + 0 + 0 + 2 = 4}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{{\bf{2 + 0 + 1 + 0 = 3}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{{\bf{2 + 0 + 1 + 2 = 5}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{{\bf{2 + 1 + 0 + 0 = 3}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{1}}&{{\bf{2 + 1 + 0 + 2 = 5}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{{\bf{2 + 1 + 1 + 0 = 4}}}&{{\bf{ > T}}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{{\bf{2 + 1 + 1 + 2 = 6}}}&{{\bf{ > T}}}&{\bf{1}}\end{array}\)

Thus, the Boolean function \({\bf{F(w,x,y,z) = w + xy + z}}\) can be represented by this threshold gate.

Hence, \({\bf{F(w,x,y,z) = w + xy + z}}\) is a threshold function.

Check whether the given is threshold or not

(h)

It is given that \({\bf{F(w,x,y,z) = wxz + x\bar y}}z\).

Let us first determine a table of values for all possible values of \(x\).

\(\begin{array}{*{20}{c}}{\bf{w}}&{\bf{x}}&{\bf{y}}&{\bf{z}}&{{\bf{F(w,x,y,z) = wxz + x\bar yz}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}\end{array}\)

Let us assign the following values:

\({w_w} = \frac{1}{2}\)(Weight of \({\bf{w}}\))

\({{\bf{w}}_{\bf{x}}}{\bf{ = }}1\)(Weight of \(x\))

\({{\bf{w}}_{\bf{y}}}{\bf{ = - }}\frac{1}{2}\)(Weight of \(y\))

\({w_z}{\bf{ = }}1\)(Weight of \(z\))

\(T = \frac{7}{4}\)

Let us then finally check that the threshold is reached for those combinations that need to return \(1\).

\(\begin{array}{*{20}{c}}{\bf{w}}&{\bf{x}}&{\bf{y}}&{\bf{z}}&{{{\bf{w}}_{\bf{x}}}{\bf{x + }}{{\bf{w}}_{\bf{y}}}{\bf{y + }}{{\bf{w}}_{\bf{z}}}{\bf{z}}}&{{\bf{Threshold}}}&{{\bf{Returnedvalues}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{{\bf{0 + 0 + 0 + 0 = 0}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{{\bf{0 + 0 + 0 + 1 = 1}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{{\bf{0 + 0 - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 0 = - }}\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{{\bf{0 + 0 - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 1 = }}\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{{\bf{0 + 1 + 0 + 0 = 1}}}&{{\bf{ < T}}}&{}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{1}}&{{\bf{0 + 1 + 0 + 1 = 2}}}&{{\bf{ > T}}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{{\bf{0 + 1 - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 0 = }}\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ < T}}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{{\bf{0 + 1 - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 1 = }}\frac{{\bf{3}}}{{\bf{2}}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 0 + 0 + 0 = }}\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 0 + 0 + 1 = }}\frac{{\bf{3}}}{{\bf{2}}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 0 - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 0 = 0}}}&{{\bf{ < T}}}&{}\\{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 0 - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 1 = 1}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 1 + 0 + 0 = }}\frac{{\bf{3}}}{{\bf{2}}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{1}}&{\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 1 + 0 + 1 = }}\frac{{\bf{5}}}{{\bf{2}}}}&{{\bf{ > T}}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 1 - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 0 = 1}}}&{{\bf{ < T}}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 1 - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + 1 = 2}}}&{{\bf{ > T}}}&{\bf{0}}\end{array}\)

Thus, the Boolean function \({\bf{F(w,x,y,z) = wxz + x\bar y}}z\) can be represented by this threshold gate.

Hence, \({\bf{F(w,x,y,z) = wxz + x\bar y}}z\) is a threshold function.