How do we conclude using De Morgan's laws that these two are equal? Question: (p∧q)→(p∨q)≡¬(p∧q)∨(p∨q)
Which steps should I take to derive the equation to the right from the equation to the left? In the book, it just shows this equation but doesn't answer how did they actually get it. Since this example in the book shows up under the De Morgan's laws section, I rightfully considered De Morgan's laws could help us to solve this problem. 
If you need the full question just tell me.
 A: Apply the definition of $a\rightarrow b$: it is 
$$(\lnot a)\vee b$$
A: Recall:  In general, $A \to B$ is equivalent to $B \lor \lnot A$.
So then in this case, $\underbrace{(p \land q)}_A \to \underbrace{(p \lor q)}_B$ is equivalent to $\underbrace{(p \lor q)}_B \lor \underbrace{\lnot (p \land q)}_{\lnot A}$
Didn't realize when I started that that was basically it..  The conclusion in the book is the same since it doesn't matter which one comes first when you use $\lor.\quad$  $B \lor \lnot A$ is the same as $\lnot A \lor B$.
A: Use a truth-table:
\begin{array}{cc|ccc|cccccc}
p & q & (p \land q) & \rightarrow & (p \lor q) & \neg & (p \land q) & \lor & (p \lor q)\\
\hline
T & T & T & \color{red}T & T & F & T & \color{red}T & T\\
T & F & F & \color{red}T & T & T & F & \color{red}T & T\\
F & T & F & \color{red}T & T & T & F & \color{red}T & T\\
F & F & F & \color{red}T & F & T & F & \color{red}T & F
\end{array}
We see that the two statementrs have the same truth-conditions (in fact, it turns out they are both tautologies), so they are equivalent.
A: If both $p$ and $q$ are true, then at least one of either $p$ or $q$ will be true.

Since this example in the book shows up under the De Morgan's laws section, I rightfully considered De Morgan's laws could help us to solve this problem. 

Yes, but it is too soon.   The step you have applies Material Implication.


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*$\qquad(p\wedge q)\to (p\vee q)~~\equiv~~\neg (p\wedge q)\vee (p\vee q)$


This sets it up to apply deMorgan's Rule next.


*

*$\qquad\phantom{(p\wedge q)\to (p\vee q)} ~~\equiv~~(\neg p\vee \neg q)\vee (p\vee q)$


Finish with association and commutation, contradiction, and then annihlation.


*

*$\qquad\phantom{(p\wedge q)\to (p\vee q)}~~\equiv~~\top$

