How to convert formula to disjunctive normal form?

Convert

$$((p \wedge q) → r) \wedge (¬(p \wedge q) → r)$$

to DNF.

This is what I've already done:

$$((p \wedge q) → r) \wedge (¬(p \wedge q) → r)$$

$$(¬(p \wedge q) \vee r) \wedge ((p \wedge q) \vee r)$$

$$((¬p \vee ¬q) \vee r) \wedge ((p \wedge q) \vee r)$$

And from this point I'm not sure how to proceed. Help would be appreciated.

Sorry, but the last line was written badly (I think). It's fixed now.

You can continue by using Distributivity of the boolean algebra:

$((¬p \vee ¬q) \vee r) \wedge ((p \wedge q) \vee r)$

$ \Leftrightarrow (¬p \vee ¬q \vee r) \wedge ((p \wedge q) \vee r)$

Here we apply distributivity:

$ \Leftrightarrow (¬p \wedge p \wedge q) \vee (¬q \wedge p \wedge q) \vee (r \wedge p \wedge q) \vee (¬p \wedge r) \vee (¬q \wedge r) \vee (r \wedge r)$

Formally, this is in disjunctive normal form now. We could further simplify:

$ \Leftrightarrow (r \wedge p \wedge q) \vee (¬p \wedge r) \vee (¬q \wedge r) \vee r$

$$((p \land q) \to r) \land (\neg(p \land q)\to r) \equiv (\neg(p \land q) \lor r) \land ((p \land q) \lor r)$$

using the distributive law

$$r \lor (\neg (p \land q) \land (p \land q)) \equiv r \lor \text{False} \equiv r$$

because $s \land \neg s \equiv \text{False}$.