Using boolean algebra, simplify $$y = \bar{s} \cdot \bar{u} + s \cdot \bar{u}+s \cdot u$$
Here is how I would solve it.
$$y=s'u' + su' + su $$
$$y=u'(s'+s)+su $$
$$y=u'+su $$
$$y=u'+s $$
Note that I use a slightly different notation, where \$\bar{s}=s' \$.
\$\rule{17cm}{0.4pt} \$
Proof that \$u'+su=u'+s \$
$$u'+su $$
The associative theorem states: \$x+yz = (x+y) \cdot (x+z) \$
$$(u'+u) \cdot (u'+s) $$
$$1 \cdot(u'+s) $$
$$u'+s $$