# I must solve this by using boolean algebra rules but i dont know how

I've got this problem there which I have no idea how to solve (by using boolean algebra rules). $$b\bar{c} + ba + \bar{a}\bar{c}$$

I have to simplify it and I don't know which rule should I use.

• Did you try anything? – Bram28 Oct 18 '17 at 20:01
• Well I'm thinking about this for about an hour but I just cant figure it out. I must must have overlooked something important. – Zdenda Oct 18 '17 at 20:07

We make use of the the distributive property and the fact that in Bollean Algebra it holds that $x \bar{x}=0$ $\forall x$.

So we have $$b\bar{c} + ba + \bar{a}\bar{c}=\\b\bar{c} + ba + \bar{a}\bar{c}+a\bar{a}=\\b(a+\bar{c})+\bar{a}(a+\bar{c})=\\(a+\bar{c})(b+\bar{a})$$

• Yes, but it doesn't match with my final result which should look like this: $$ab\ + \bar{a}\bar{c}\\$$ – Zdenda Oct 18 '17 at 20:40
• @Zdenda Perhaps you should have bothered writing the result you wish to reach in the first place. Ok, can you make a table of all possible values of $a,b,c$? That will answer your question. In the meantime I will try to see how we can go there by exploiting basic properties. – MathematicianByMistake Oct 18 '17 at 20:48

The Consensus Theorem says that:

Consensus

$$pq + \bar q r + p r = p q + \bar q r$$

With Consensus, you can simplify your expression in 1 step:

$$b\bar{c} + ba + \bar{a}\bar{c} = \text{ (Consensus)}$$

$$ba + \bar{a}\bar{c}$$

.. but you probably weren't given Consensus. So, let's prove Consensus:

Consensus

$$pq + \bar q r + p r = p q + \bar q r$$

Proof:

$$p q + \bar q r + p r = \text{ (Adjacency)}$$

$$p q + \bar q r + p q r + p \bar q r = \text{ (Absorption)}$$

$$p q + \bar q r$$

And if you don't have Adjacency or Absorption:

$$p = p q + p \bar q$$

Proof:

$$p = \text{ (Identity)}$$

$$p 1 = \text{ (Complement)}$$

$$p (q + \bar q) = \text{ (Distribution)}$$

$$p q + p \bar q$$

Absorption

$$p + p q = p$$

Proof:

$$p + p q = \text{ (Identity)}$$

$$p 1 + p q = \text{ (Distribution)}$$

$$p ( 1 + q )= \text{ (Annihilation)}$$

$$p 1 = \text{ (Identity)}$$

$$p$$