Can I simplify: $(¬P ∧ Q) ∨ (P ∧ ¬Q)$? I got stuck on this development:  
$$\begin{align} (¬P ∧ Q) ∨ (P ∧ ¬Q) & \iff ((¬P ∧ Q) ∨ P) ∧ ((¬P ∧ Q) ∨ ¬Q) \tag{1} \\  
&\iff (P ∨ Q) ∧ (¬P ∨ ¬Q)   \tag{2}  \\
\end{align}$$
Can't this also be written as this using the associative and tautology Rules?
$$\begin{align} (¬P ∧ Q) ∨ (P ∧ ¬Q) & \iff ((¬P ∧ Q) ∨ P) ∧ ((¬P ∧ Q) ∨ ¬Q) \tag{1'} \\  
&\iff (Q ∧ (¬P ∨ P)) ∧ (¬P ∧ (Q ∨ ¬Q))   \tag{2'}  \\
&\iff (Q ∧ ¬P)   \tag{2'}  \\
\end{align}$$
Is this correct?
 A: $(\neg P\land Q)\lor(P\land\neg Q)$ is not equivalent to $Q\land \neg P$ -- for example when $Q$ is false and $P$ is true, the former is true and the latter is false.

It looks like you've been fooled into thinking that $(A\land B)\lor C$ is the same as $A\land(B\lor C)$. This is not the case. For example, when $C$ is true but $A$ and $B$ are false, $(A\land B)\lor C$ will be true but $A\land(B\lor C)$ will be false.
$\land$ is associative by itself, and $\lor$ is associative by itself, but they don't associate with each other.
A: The statement means: “$P$ or $Q$, the ‘or’ being exclusive,” or “exactly one of $P$ and $Q$.” This is the same as “$P$ or $Q$, but not both:” $$(P\vee Q)\wedge\lnot (P\wedge Q).$$
A: In this case, you can tell directly by looking at the form of the formula in question that it cannot be simplified in the way you're looking for: it is already "maximally simple" in the sense that it is in disjunctive normal form. A propositional formula with k variables $P_1, \ldots, P_k$ has a unique representation as a disjunction of 0 or more subformulas of the form $(\neg)^{\pm 1}P_1 \wedge \cdots \wedge (\neg)^{\pm 1}P_k$, where $(\neg)^{\pm 1}P$ can either be $P$ or $\neg P$.
