Reduction of Boolean expression

Question

Reduce the following Boolean expression: $q’ (r’s + rs) + qrs$.

$$\begin{align} q' (r's + rs) + qrs\quad&\\ q’r’s + q’rs + qrs\quad&\text{distributive law}\\ q’ (r’s + rs) + qrs\quad&\\ s (q’r’ + q’r + qr)\quad &\text{distributive law}\\ s (q’ (r’ + r) + qr)\quad&\text{distributive law}\\ s (q’1 + qr)\quad&\text{inverse law}\\ s(q’ + qr)\quad&\text{identity law}\\ sq' + sqr\quad&\text{distributive law} \end{align}$$ Is there any way to reduce this further? Since absorption law doesn't apply I haven't found any other laws that could reduce this expression anymore. Please help.

Answer 1

Hint.

$$ r's+rs=s \Rightarrow q'(r's+rs) = q's $$

Answer 2

Yes, you can reduce it further. Let's start one line before your last line:

$$\begin{align} s(q’ + qr)\quad&\text{identity law}\\ s(q' + q)(q'+r)\quad&\text{distributive law}\\ s1(q'+r)\quad&\text{inverse law}\\ s(q'+r)\quad&\text{identity law}\\ sq'+sr\quad&\text{distributive law}\\ \end{align}$$