Question

Show that \({\bf{x\bar y + y\bar z + \bar xz = \bar xy + \bar yz + x\bar z}}\).

Short answer

The given \(x\bar y + y\bar z + \bar xz = \bar xy + \bar yz + x\bar z\) is proved.

Step-by-step solution

Definition

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

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

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

Using the Boolean product and sum on the left-hand side

You first determine the table representing the left side of the equation.

\(F(x,y,z) = x\bar y + y\bar z + \bar xz\)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

\(\begin{array}{*{20}{r}}x&{ y}&{ z}&{ \bar x}&{\bar xz}&{ \bar y}&{ x\bar y}&{ \bar z}&{ y\bar z}&{ xy + xz + yz}\\0&0&0&1&0&1&0&1&0&0\\0&0&1&1&1&1&0&0&0&1\\0&1&0&1&0&0&0&1&1&1\\0&1&1&1&1&0&0&0&0&1\\1&0&0&0&0&1&1&1&0&1\\1&0&1&0&0&1&1&0&0&1\\1&1&0&0&0&0&0&1&1&1\\1&1&1&0&0&0&0&0&0&0\end{array}\)

Using the Boolean product and sum on the right-hand side

Next, you determine the table representing the right side of the equation.

\(F(x,y,z) = \bar xy + \bar yz + x\bar z\)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

\(\begin{array}{*{20}{c}}x&{ y}&{ z}&{ \bar x}&{ \bar xy}&{ \bar y}&{ \bar yz}&{ \bar z}&{ x\bar z}&{ \bar xy + \bar yz + x\bar z}\\0&0&0&1&0&1&0&1&0&0\\0&0&1&1&0&1&1&0&0&1\\0&1&0&1&1&0&0&1&0&1\\0&1&1&1&1&0&0&0&0&1\\1&0&0&0&0&1&0&1&1&1\\1&0&1&0&0&1&1&0&0&1\\1&1&0&0&0&0&0&1&1&0\\1&1&1&0&0&0&0&0&0&1\end{array}\)

You obtain that the last column in the two tables is identical.

Therefore, you get \(x\bar y + y\bar z + \bar xz = \bar xy + \bar yz + x\bar z\).