Logic: Can you drop parentheses in a conjunction? In propositional logic, $p \land (q \land r) = (p \land q) \land r$ , where $p, q$ and $r$ are propositions.
Does this mean $p \land (q \land r) = p \land q \land r$ ? If so, why?
 A: Yes, you are right: conjunction is associative. Thus, it doesn't matter where you put the parentheses. Hence, when you see "$p \land q \land r$", it means that you may insert the parentheses anywhere and you will get the same result.
A: I highly recommend avoiding "p∧q∧r" and the like when talking in a proof-theoretic context.  In a proof-theoretic context a proof is a sequence of well-formed formulas (wffs).  I can tell which wff p∧(q∧r) stands for and I can tell which wff (p∧q)∧r stands for (neither of those strings are wffs).  But, I can't tell which wff p∧q∧r stands for.  It could mean (p∧(q∧r)) or ((p∧q)∧r) or it could mean both (p∧(q∧r)) and ((p∧q)∧r).  And thus, in a proof-theoretic context p∧q∧r is meaningless.
A: Informally, one might say that if three things are true ($p \land q \land r$), then also one thing and two things — $p \land (q \land r)$ — and two things and one thing — $(p \land q) \land r$ — are true.
In a formal deductive system commonplace in symbolic logic, we want to be more formal. We usually deduce consequences from a set of premises. For example we deduce $p \land q$ from the set $\{p, q\}$. This is written:
$$\{p,q\} \vdash p \land q$$
where the braces are usually omitted, and $\vdash$ is voiced proves. Similarly:
$$p \land q \vdash p \qquad\qquad p\land q \vdash q$$
(where the premises thus are considered singleton sets) because this is simply throwing information away. 
If we now allow chains of deduction, then, one easily obtains:
$$\begin{align*}\{p,q,r\} &\vdash p \land q& \{p \land q, r\} &\vdash (p \land q) \land r\\
\{p,q,r\}&\vdash q \land r&\{p, q \land r\}&\vdash p \land(q \land r)
\end{align*}$$
and from the rightmost conclusion the so-called $\land$-elimination rules  easily allow deriving $p, q$ and $r$ again. Thus either of the parenthesized expressions corresponds exactly to the truth of the three letters $p,q,r$. Thus we say they're (provably) equivalent:
$$p \land (q \land r) \dashv\vdash (p \land q) \land r$$
Thus we have demonstrated that the following definitions of $p \land q \land r$ are all equally sensible:


*

*As $p \land (q \land r)$;

*As $(p \land q) \land r$;

*By stating $p \land q \land r$ is true precisely when $p,q,r$ are all true.

