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.
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.
On
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:
Adjacency
$$p = p q + p \bar q$$
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:
$$p + p q = \text{ (Identity)}$$
$$p 1 + p q = \text{ (Distribution)}$$
$$p ( 1 + q )= \text{ (Annihilation)}$$
$$p 1 = \text{ (Identity)}$$
$$p$$
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 $$b\bar{c} + ba + \bar{a}\bar{c}=\\b\bar{c} + ba + \bar{a}\bar{c}+a\bar{a}=\\b(a+\bar{c})+\bar{a}(a+\bar{c})=\\(a+\bar{c})(b+\bar{a})$$