Question

truth table 1.(x+z)' . (y.y') + (x+y')'​

Answer 1

$\bf\huge\underline{Answer}:$

Truth Table of (x+z)' . (y.y') + (x+y')' :

$\begin{gathered}\boxed{\begin{array}{c|c|c|c|c|c|c|c|c}\bf{x} & \bf{y} & \bf{z} & \bf{y'} & \bf{x+z} & \bf{x+y'} & \bf{(x+z)'} & \bf{y.y'} & \bf{(x+y')'}\\ \sf{0} & \sf{0} & \sf{0} & \sf{1} & \sf{0} & \sf{1} & \sf{1} & \sf{0} & \sf{0} \\ \sf{0} & \sf{0} & \sf{1} & \sf{1} & \sf{1} & \sf{1} & \sf{0} & \sf{0} & \sf{0} \\ \sf{0} & \sf{1} & \sf{0} & \sf{0} & \sf{0} & \sf{0} & \sf{1} & \sf{0} & \sf{1} \\ \sf{0} & \sf{1} & \sf{1} & \sf{0} & \sf{1} & \sf{0} & \sf{0} & \sf{0} & \sf{1} \\ \sf{1} & \sf{0} & \sf{0} & \sf{1} & \sf{1} & \sf{1} & \sf{0} & \sf{0} & \sf{0} \\ \sf{1} & \sf{0} & \sf{1} & \sf{1} & \sf{1} & \sf{1} & \sf{0} & \sf{0} & \sf{0} \\ \sf{1} & \sf{1} & \sf{0} & \sf{0} & \sf{1} & \sf{1} & \sf{0} & \sf{0} & \sf{0} \\ \sf{1} & \sf{1} & \sf{1} & \sf{0} & \sf{1} & \sf{1} & \sf{0} & \sf{0} & \sf{0}\end{array}}\end{gathered}$

Knowledge Bytes :

• The result of any logical statement are termed as truth tables and their representation is shown in a table called truth table.

Logical NOT :

• This operator provides the complement of the given binary valued quantity.

$\begin{gathered}\boxed{\begin{array}{c|c}\bf{A} & \bf{\bar{A}} \\ \sf{0} & \sf{1} \\ \sf{1} & \sf{0}\end{array}}\end{gathered}$

Logical OR :

• This operator results in logical addition of binary valued quantities.

$\begin{gathered}\boxed{\begin{array}{c|c|c}\bf{A} & \bf{B} & \bf{A+B}\\ \sf{0} & \sf{0} & \sf{0} \\ \sf{0} & \sf{1} & \sf{1}\\ \sf{1} & \sf{0} & \sf{1}\\ \sf{1} & \sf{1} & \sf{1}\end{array}}\end{gathered}$

Logical AND :

• This operator results in the product of two or more binary valued quantities.

$\begin{gathered}\boxed{\begin{array}{c|c|c}\bf{A} & \bf{B} & \bf{A.B}\\ \sf{0} & \sf{0} & \sf{0} \\ \sf{0} & \sf{1} & \sf{0}\\ \sf{1} & \sf{0} & \sf{0}\\ \sf{1} & \sf{1} & \sf{1}\end{array}}\end{gathered}$