# Using boolean algebra, simplify $$y = \bar{s} \cdot \bar{u} + s \cdot \bar{u}+s \cdot u$$

I have the following function, that I want to minimise using boolean algebra:

$$y = \bar{s} \cdot \bar{u} + s \cdot \bar{u}+s \cdot u$$

Here's my attempt:

$$\bar{s} \cdot \bar{u} + s \cdot \bar{u}+s \cdot u = \bar{u} \cdot (\bar{s} +s)+s \cdot u = \bar{u} \cdot 1+s \cdot u=\bar{u} + s \cdot u = \bar{u} +s$$

In the last step I used the absorption property but I was wondering if there is another way to solve:

$$\bar{u} + s \cdot u$$

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$$

• Ah. It's so clear now. You simply use the associative theorem backwards. Thank you so much! Commented Sep 9, 2020 at 14:09

In boolean algebra you can duplicate a term without altering the final result, sometimes this makes things easier to simplify.

$$y = \bar{s} \cdot \bar{u} + s \cdot \bar{u}+s \cdot u$$

We notice that $$\s \cdot \bar{u}\$$ can be combined with both of the other two terms to produce simpler terms, so we duplicate it.

$$y = \bar{s} \cdot \bar{u} + s \cdot \bar{u}+ s \cdot \bar{u}+s \cdot u$$

$$y = (\bar{s} + s) \cdot \bar{u}+ s \cdot (\bar{u}+ u)$$

$$y = \bar{u}+ s$$