Problem with simplification in discrete math I am doing my homework in discrete mathematics and I need your help..
I can' t find the way how to simplify this equation. I had to get Minimal Disjunctive Normal Form by just simplifying minimal conjunctive normal form. I am 100 % sure these two - Minimal Disjunctive Normal Form and minimal conjunctive normal form - are correct so there are no mistakes there. And the truth tables of minimal disjunctive normal form and the DNK i am stuck at match, so I should be able to simplify it.
I have to get this: 
$$x_1x_4 \lor x_1x_2x_3\lor ((\neg x_1)x_3(\neg x_4))$$
out of this: 
$$(x_1 \lor (\neg x_4))(x_3 \lor x_4)((\neg x_1 ) \lor x_2 \lor x_4)$$
But I'm stuck at this particular point: 
$$x_1x_2x_3 \lor x_1x_4 \lor (\neg x_1)x_3(\neg x_4) \lor x_2x_3(\neg x_4)$$
 A: So somehow you have to show that the disjunct $x_2x_3(\neg x_4)$ is already covered by the expression $x_1x_4 \lor x_1x_2x_3\lor ((\neg x_1)x_3(\neg x_4))$. Since $x_2x_3(\neg x_4)$ is true only when $x_4$ is false, the $x_1x_4$ term won’t come into play, and we should try to get $x_2x_3(\neg x_4)$ out of $x_1x_2x_3\lor ((\neg x_1)x_3(\neg x_4))$. This involves application of absorption and of the fact that we can insert or remove $x\lor(\neg x)$ at will. I’ve used color coding below to show how some of the less obvious steps work.
$$\begin{align*}
\color{blue}{x_1x_2x_3}\lor\color{green}{\big((\neg x_1)x_3(\neg x_4)\big)}&\equiv\color{blue}{x_1x_2x_3\lor x_1x_2x_3(\neg x_4)}\lor\color{green}{\big((\neg x_1)\big(x_2\lor(\neg x_2)\big)x_3(\neg x_4)\big)}\\
&\qquad\color{green}{\lor\big((\neg x_1)x_3(\neg x_4)\big)}\\
&\equiv x_1x_2x_3\lor\color{brown}{x_1x_2x_3(\neg x_4)\lor(\neg x_1)x_2x_3(\neg x_4)}\\
&\qquad\lor(\neg x_1)(\neg x_2)x_3(\neg x_4)\lor\big((\neg x_1)x_3(\neg x_4)\big)\\
&\equiv x_1x_2x_3\lor\color{brown}{\big(x_1\lor(\neg x_1)\big)x_2x_3(\neg x_4)}\\
&\qquad\lor(\neg x_1)(\neg x_2)x_3(\neg x_4)\lor\big((\neg x_1)x_3(\neg x_4)\big)\\
&\equiv x_1x_2x_3\lor x_2x_3(\neg x_4)\\
&\qquad\color{blue}{\lor(\neg x_1)(\neg x_2)x_3(\neg x_4)\lor\big((\neg x_1)x_3(\neg x_4)\big)}\\
&\equiv x_1x_2x_3\lor x_2x_3(\neg x_4)\lor\color{blue}{\big((\neg x_1)x_3(\neg x_4)\big)}
\end{align*}$$
