Is this expression $(a \land b) \land c$ the same as $a \land b \land c$? I'm trying to create a Conjunctive Normal Form, which is from $(a \land b) \leftrightarrow c$.
Now the issue is that I'm not sure whether $(a \land b) \land c$ is the same as saying $a \land b \land c$.
Especially when I will be applying De Morgan's Laws, so in that case whether $\neg(a \land b) \land \neg c$ is the same as $\neg a \land \neg b \land \neg c$.
Thanks.
 A: $$(A\land B)\land C=A\land(B\land C)=A\land B\land C.$$
But
$$\lnot(A\land B)\land \lnot C\ne \lnot A\land \lnot B\land\lnot C.$$
A: I clarify a bit better : because of the conjuction is a binary operator, then you have to choose which of the operations you want to do first. So , suppose $A\land B\land C$ stays for $(A\land B)\land C$. But you can prove $((A\land B)\land C) \iff (A\land (B\land C))$ because the conjuction is associative. So :
$$((A\land B)\land C) \iff (A\land (B\land C)) \iff (A\land B\land C)$$
A: The symbol $\land$ is a binary operation. It takes two well-formed formulae, $\varphi$ and $\psi$, and produces a third, $\varphi \land\psi$. It is associative, in the sense that, for any three well-formed formulae, $\theta$, $\varphi$, and $\psi$, we have
$$\theta\land(\varphi \land \psi)\equiv (\theta\land\varphi)\land\psi.\tag{$A$}$$
(Can you prove $(A)$?)

 A proof tree, generated here, is 

Now it's just a matter of the generalised associativity law; the point being:
$$A\land B\land C$$
means, technically, both $(A\land B)\land C$ and $A\land(B\land C)$, since they are equivalent; so the parentheses are still there - one just omits them for convenience.
A: 
These are the same laws in logic, just replace the notations $\cup=\lor$ and $\cap=\land$ and $\bar X=\lnot x$
So these equalities are true in particular
$\overline{(A\land B)}\land \bar C=(\bar A\lor \bar B)\land \bar C=(\bar A\land\bar C)\lor(\bar B\land \bar C)=\overline{(A\lor C)}\lor\overline{(B\lor C)}$
