By writing the function in disjunctive normal form, design an equivalent system of switches, that is one having an equal switching function.
$ f(x_1,x_2,x_3) = x_1(x_2(x_1 \oplus x_3) \oplus x_3 \overline{x_2}) $
I've had a look in my book and there is very little mention of disjunctive normal form. I've had a look online and I really have no idea to go converting this function into it. Any help would be great.
Thanks
Let us fix some notation first. We use $\phi_1 \wedge \phi_2$ to denote "$\phi_1$ and $\phi_2$" (this is called a conjunction), $\phi_1 \vee \phi_2$ to denote "$\phi_1$ or $\phi_2$" (this is called a disjunction and $\neg \phi_1$ to denote not $\phi_1$ (this is called a negation). We use $x_1, x_2 \ldots$ to denote Boolean variables and abbreviate $\neg x$ as $\overline{x}$. A literal $\ell$ is either a Boolean variable $x$ or a negated Boolean variable $\overline{x}$.
A Boolean formula $\phi$ is a disjunctive normal form if $\phi$ is of the form
$$\bigvee^{m}_{i=1} \bigwedge^{m_i}_{j=1} \ell_{ij}$$
A Boolean formula $\phi$ is a conjunctive normal form if $\phi$ is of the form
$$\bigwedge^{m}_{i=1} \bigvee^{m_i}_{j=1} \ell_{ij}$$
That is, $\phi$ is a conjunction of disjunctions of literals.
A central theorem is that for every Boolean formula $\phi$ we can construct an equivalent formula in disjunctive normal form and an equivalent formula in conjuctive normal form.
To do this, use the identities
$$\neg (\phi_1 \vee \phi_2) \equiv \neg \phi_1 \wedge \neg \phi_2$$ $$\neg (\phi_1 \wedge \phi_2) \equiv \neg \phi_1 \vee \neg \phi_2$$ $$\phi_1 \vee (\phi_2 \wedge \phi_2) \equiv (\phi_1 \vee \phi_2) \wedge (\phi_1 \vee \phi_3)$$ $$\phi_1 \wedge (\phi_2 \vee \phi_2) \equiv (\phi_1 \wedge \phi_2) \vee (\phi_1 \wedge \phi_3)$$
to push the negations inwards and to group the disjunctions together and group the conjuctions together.
A worked example may be helpful here. Watch https://www.youtube.com/watch?v=D7nyonAsIo8 .