In the following logic circuit the sequence of the inputs A ,   B are ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 )…

Look at the above circuit, the value of $P$ can be found by the idea of AND gate, $P=A·B$ and the value of $Q$, can be calculated by the idea of OR gate, $Q=A+B$. Then expression for $Y$ can be taken by the idea of NAND gate, $\Rightarrow Y=\overline{P·Q}$ by the idea of Demorgan's theorem, $Y=\overline{P}+\overline{Q}=\overline{A}+\overline{B}+\overline{A}·\overline{B}$

 

$\begin{array}{ccc}\mathrm{A}& \mathrm{B}& \mathrm{Y}=\overline{\mathrm{A}}+\overline{\mathrm{B}}+\overline{\mathrm{A}}·\overline{\mathrm{B}}\\ 0& 0& 1\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}$