show $((p \to q) \land (r \to s) \land (p \lor r )) \to (q \lor s)$ is a tautology I have tried the following but then got stuck:
I apply the implication rule to the first two elements:
$((\lnot p \lor q) \land (\lnot r \lor s) \land (p \lor r)) \to (q \lor s)$
Then again to the entire equation:
$\lnot((\lnot p \lor q) \land (\lnot r \lor s) \land (p \lor r)) \lor (q \lor s)$
I am not sure how to distribute the negation sign.

De Morgan's Laws:
$\lnot (p \lor q) = \lnot p \land \lnot q$ (note that the "or" sign changes into "and")
$\lnot (p \land q) = \lnot p \lor \lnot q$ (note that the "and" sign changes into "or")
This is because the negation of "and" is "or" and vice versa. 
So from your situation:
\begin{align}
&\lnot((\lnot p \lor q) \land (\lnot r \lor s) \land (p \lor r)) \lor (q \lor s)
\\
\equiv\,& (p \land \lnot q) \lor (r \land \lnot s) \lor (\lnot p \land \lnot r) \lor (q \lor s)
\end{align}
From here you can expand everything out, use some associative properties, and voila! 
