logic distributivity with only conjunction (or disjunction) Hi I read real analysis book, but some part strikes me weird.. so far I knew that 
$$((p\wedge q)\wedge r) \Leftrightarrow (p\wedge (q\wedge r))$$
but the textbook said that 
$$((p\wedge q)\wedge r) \Leftrightarrow (p \wedge r) \wedge (q \wedge r)$$ 
and
$$(p\wedge (q\wedge r)) \Leftrightarrow (p \wedge q) \wedge (p \wedge r)$$
which means that 
$$(p \wedge q) \wedge (p \wedge r)\Leftrightarrow (p \wedge r) \wedge (q \wedge r)$$ 
I guess this only makes sense when we can ignore parenthesis such that 
$$(p \wedge q) \wedge (p \wedge r) \Leftrightarrow p \wedge q \wedge p \wedge r \Leftrightarrow  q \wedge p \wedge r $$ as $ p \wedge p \Leftrightarrow p$
Am I doing the right thing or do I omit some important part? If, like I said, we can omit parenthesis when statements are connected with only conjunction or disjunction, what is the point of using parenthesis?
 A: It depends on the relevant standard for "doing the right thing".  
Semantically, an expression of the form 
$\Phi(p,q,\ldots)\Leftrightarrow \Psi(p,q,\ldots)$ 
means that the formulas on the left and right sides of $\Leftrightarrow$ are either both true or both false, no matter how truth and falsehood are distributed amongst $p,q,\ldots$.
Now, you can think of $\land$ as a function which returns the minimum of the values of its arguments, where falsehood is smaller than truth.  Suppose you have some function $\Phi$ which is defined by composition from $\land$.  Then all that matters to the value of $\Phi$ is which of $p,q,\ldots$ appear as its arguments.  So, the truth or falsehood of $\Phi(p,q,\ldots)$ is unaffected by dropping parentheses, reordering of arguments, and adding and deleting copies of the same arguments.
On the other land, logical relationships get more complicated when you go beyond $\land$ into other truth-functional connectives and then the quantifiers.  Then you may want to get a more fine-grained approach by considering a system of proof: this lays out a collection of atomic steps which can be combined together to yield demonstrations of the dependencies between truth-values of formulas.  Specifically, a proof demonstrates that one formula is a consequence of some others, in the sense that the first must be true in any logically possible world at which all the others are true.
There are a couple of rules which suffice to demonstrate all the logical relationships between formulas built up just from $\land$:


*

*from $\Phi$ and $\Psi$ together, infer $\Phi\land \Psi$

*from $\Phi\land \Psi$, infer $\Phi$ or infer $\Psi$.


To establish a conclusion of the form $\Phi\Leftrightarrow \Psi$ using these rules, you need to break it in half, arguing for $\Phi\Rightarrow \Psi$ and $\Psi\Rightarrow \Phi$.
Finally, to go back to the question you raised at the end.  Traditionally, functions are conceived of as having a fixed "arity" or number of argument-places.  In particular, conjunction is regarded as having arity two.  So when conjunction is regarded as a two-place function in the traditional sense, then $p\land q\land r$ is ambiguous between $(p\land q)\land r$ and $p\land(q\land r)$, and it's a substantial (if obvious) property of the function $\land$ that the two formulas are semantically equivalent.  
A: You've already got an excellent answer, but simply put it is precisely because $\land$ (which you can define by its truth-table) is associative, and hence you can write a conjunction of multiple propositions without brackets, since all groupings give the same result by associativity.
Also, one point that was not mentioned in the other answer is that $p \land p$ is equivalent to $p$, and hence having duplicates in a conjunction of propositions makes no difference to the truth value.
A: Edit.
The point is that you need to define formally the expression
$$
p∧q∧p∧r
$$
In your case, you can only use it once you have proved that whatever way you put parentheses in this expression, say $(p∧q)∧(p∧r)$ or $((p∧q)∧p)∧r$ or $p∧(q∧(p∧r))$, you obtain the same result.
To convince you of the necessity of proving such a result, just think of an expression like
$$
p-q-p-r
$$
As you can see $(p-q)-(p-r)= r-q$, but $((p-q)-p)-r = -q -r$ and $p-(q-(p-r)) = 2p-(q+r)$. Thus your argument is based on the associativity and the idempotence of the operation $\wedge$, two things that need to be formally proved as a prerequisite.
