Simplify the Boolean function \((x \cdot y)+\left[\left(x+y^{\prime}\right) \cdot y\right]^{\prime}\)

\((x \cdot y)+\left[\left(x+y^{\prime}\right) \cdot y\right]^{\prime} \)
\( =(x \cdot y)+\left[\left(x+y^{\prime}\right)^{\prime}+y^{\prime}\right] {\qquad \left[\because(a \cdot b)^{\prime}=a^{\prime}+b^{\prime}\right]} \)
\( =(x \cdot y)+\left[x^{\prime} \cdot\left(y^{\prime}\right)^{\prime}+y^{\prime}\right] ~{\qquad\left[\because(a+b)^{\prime}=a^{\prime} \cdot b^{\prime}\right]} \)
\( =(x \cdot y)+\left[x^{\prime} \cdot y+y^{\prime}\right] {\qquad\quad~~\left[\because\left(a^{\prime}\right)^{\prime}=a\right]} \)
\( =x \cdot y+y^{\prime}+x^{\prime} \cdot y {\qquad\qquad\quad[\because a+b=b+a]} \)
\( =x \cdot y+\left(y^{\prime}+x^{\prime}\right) \cdot\left(y^{\prime}+y\right)\quad~~[\text { by distributive law }] \)
\( =x \cdot y+\left(y^{\prime}+x^{\prime}\right) \cdot 1 ~{\qquad\qquad\left[\because a+a^{\prime}=1\right]} \)
\( =x \cdot y+x^{\prime}+y^{\prime} \)
\( =x \cdot y+(x \cdot y)^{\prime} {\qquad\qquad\qquad\left[\because a^{\prime}+b^{\prime}=(a \cdot b)^{\prime}\right]} \)
\( =1 ~{\qquad\qquad\qquad\qquad\qquad\quad\left[\because a+a^{\prime}=1\right]}\)