Don't understand the following equivalence after distributivity for $(\neg p\vee \neg q)\wedge(p\vee q)$

Question

In my logic course I saw the following equivalence, but I don't understand how they get there.$(\neg p\vee \neg q)\wedge(p\vee q)\, \Longleftrightarrow \, (\neg p\wedge p)\vee (\neg p\wedge q)\vee(\neg q\wedge p)\vee(\neg q\wedge q)$

I can see it's equivalent by setting up a truth table, but I just don't understand how they used the logical equivalence for distributivity which is $\varphi \vee (\psi \wedge \chi )\Longleftrightarrow (\varphi \vee \psi )\wedge (\varphi \vee \chi )$ to reach the formula on the right side.

Thanks in advance!

Answer 1

First step: we have to distribute $(p∨q)$ over $(¬p∨¬q)$ to get:

$[(p∨q) \land \lnot p] \lor [(p∨q) \land \lnot q]$.

Second setp: apply Distributivity again twice to get:

$(\lnot p \land p) \lor (\lnot p \land q) \lor (p \land \lnot q) \lor (q \land \lnot q)$.

Finally, we may use $(\lnot p \land p) \equiv (q \land \lnot q) \equiv \text {FALSE }$ and $(\text { FALSE} \lor \alpha) \equiv \alpha$ for a formula $\alpha$ whatever to get the equivalent:

$(\lnot p \land q) \lor (p \land \lnot q)$.

Answer 2

They used it twice, once the left distributivity, then again the right distributivity.

$$\begin{array}{rcl}(p\lor q)\land(r\lor s)&\Leftrightarrow &((p\lor q)\land r)\lor((p\lor q)\land s)\\&\Leftrightarrow &(p\land r)\lor (q\land r)\lor(p\land s)\lor(q\land s)\end{array}$$

This is no different from using identities such as $(a+b)(c+d)=ac+ad+bc+bd$ in arithmetic.