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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:

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

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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) \$ $$(u'+u) \cdot (u'+s) $$ $$1 \cdot(u'+s) $$ $$u'+s $$

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  • \$\begingroup\$ Ah. It's so clear now. You simply use the associative theorem backwards. Thank you so much! \$\endgroup\$
    – Ski Mask
    Sep 9 '20 at 14:09
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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.

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

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

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

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