DNF form ( $ \neg ((p \wedge q) \equiv (r \vee s)) $ ) I need your help bringing $ \neg ((p \wedge q) \equiv (r \vee s)) $ into DNF form (Disjunctive normal form). I have tried many times but appear to get slightly different answers each time. 
Wolfram alpha gives an answer but unfortunately no steps on how to reach this answer. I have not been able to reach this answer in my attempts. Here is a link to wolfram alpha's solution: https://www.wolframalpha.com/input/?i=~(+(((p+%26%26+q)+%3C--%3E+(r+%7C%7C+s)))+)
Thank you very much! 
:)
 A: The truth table of $p\land q$ is:
$$
\begin{array}{c|c|c}
p&q&p\land q\\\hline
0&0&0\\\hline
0&1&0\\\hline
1&0&0\\\hline
1&1&1\\\hline
\end{array}
$$
The truth table of $r\lor s$ is:
$$
\begin{array}{c|c|c}
r&s&r\lor s\\\hline
0&0&0\\\hline
0&1&1\\\hline
1&0&1\\\hline
1&1&1\\\hline
\end{array}
$$
The truth table of $t\equiv u$ is:
$$
\begin{array}{c|c|c}
t&u&t\equiv u\\\hline
0&0&1\\\hline
0&1&0\\\hline
1&0&0\\\hline
1&1&1\\\hline
\end{array}
$$
Putting all together, the truth table of $\neg((p\land q)\equiv (r\lor s))$ is:
$$
\begin{array}{l|c|c|c|c|}
{}_{\Large{rs}}~~{}^{\Large{pq}}&00&01&11&10\\\hline
00&0&0&1&0\\\hline
01&1&1&0&1\\\hline
11&1&1&0&1\\\hline
10&1&1&0&1\\\hline
\end{array}
$$
You can now see all the minterms that must be OR'ed in order to give the DNF:
The minterm associated to the isolated $1$ under $(p,q,r,s)=(1,1,0,0)$ that is thus given by $p\land q\land \neg r\land\neg s$
The minterm associated to:
$$
\begin{array}{l|c|c|c|c|}
{}_{\Large{rs}}~~{}^{\Large{pq}}&00&01\\\hline
01&1&1\\\hline
11&1&1\\\hline
\end{array}
$$
is given by $\neg p\land s$.
The minterm associated to:
$$
\begin{array}{l|c|c|c|c|}
{}_{\Large{rs}}~~{}^{\Large{pq}}&00&01\\\hline
11&1&1\\\hline
10&1&1\\\hline
\end{array}
$$
is given by $\neg p\land r$
The minterm associated to:
$$
\begin{array}{l|c|c|c|c|}
{}_{\Large{rs}}~~{}^{\Large{pq}}&00&10\\\hline
01&1&1\\\hline
11&1&1\\\hline
\end{array}
$$
is given by $\neg q\land s$
The minterm associated to:
$$
\begin{array}{l|c|c|c|c|}
{}_{\Large{rs}}~~{}^{\Large{pq}}&00&10\\\hline
11&1&1\\\hline
10&1&1\\\hline
\end{array}
$$
is given by $\neg q\land r$.
The OR of all of them gives:
$$\neg((p\land q)\equiv (r\lor s))=(p\land q\land \neg r\land\neg s)\lor(\neg p\land s)\lor(\neg p\land r)\lor(\neg q\land s)\lor(\neg q\land r)$$
A: You can do all of this algebraicaly using the following algorithm:
$$ \neg ((p \wedge q) \equiv (r \vee s))$$
Rewrite all $\leftrightarrow$'s ($\equiv$'s) using $P \leftrightarrow Q \Leftrightarrow ((P \rightarrow Q) \land (Q \rightarrow P))$
$$\neg (((p \land q) \rightarrow (r \lor s)) \land ((r \lor s) \rightarrow (p \land q)))$$
Rewrite all $\rightarrow$'s using $P \rightarrow Q \Leftrightarrow \neg P \lor Q$:
$$\neg ((\neg (p \land q) \lor (r \lor s)) \land (\neg (r \lor s) \lor (p \land q)))$$
Keep applying DeMorgan until all negations are 'inside', removing any double negations as they may appear:
$$\neg (\neg (p \land q) \lor (r \lor s)) \lor \neg (\neg (r \lor s) \lor (p \land q))$$
$$ (\neg \neg (p \land q) \land \neg (r \lor s)) \lor  (\neg \neg (r \lor s) \land \neg (p \land q))$$
$$ ((p \land q) \land (\neg r \land \neg s)) \lor  ((r \lor s) \land (\neg p \lor \neg q))$$
Keep distributing any $\land$'s over $\lor$'s until it is in DNF:
$$ ((p \land q) \land (\neg r \land \neg s)) \lor  (((r \lor s) \land \neg p) \lor ((r \lor s) \land \neg q))$$
$$ ((p \land q) \land (\neg r \land \neg s)) \lor  ((r \land \neg p) \lor (s \land \neg p)) \lor ((r \land \neg q) \lor (s \land \neg q))$$
(Drop unnecessary parentheses:)
$$ (p \land q \land \neg r \land \neg s) \lor  (r \land \neg p) \lor (s \land \neg p) \lor (r \land \neg q) \lor (s \land \neg q)$$
