# Show that $(p \land q) \to (p \lor q)\equiv(p \lor ¬p) \land (q \lor \neg q)$

I'm stuck while trying to show that $$(p \land q) \to (p \lor q)$$ is logically equivalent to $$(p \lor ¬p) \land (q \lor \neg q)$$. I know they're equivalent (they're tautologies) because they both have the same truth table but I'm not sure how to start from $$(p \land q) \to (p \lor q)$$ and finally get $$(p \lor ¬p) \land (q \lor \neg q)$$.

What I have done so far is this:

\begin{align*} (p \land q) \to (p \lor q)&\equiv\neg(p \land q) \lor (p \lor q)&\text{(Definition of \to)}\\ &\equiv(\neg p\lor\neg q)\lor (p \lor q)&\text{(De Morgan's Law)}\\ &\equiv(\neg p\lor\neg p)\lor (q \lor q)&\text{(Associativity and commutativity of \lor)}\\ \end{align*}

I don't see how to get a conjunction from there. I know I could say both disjunctions on my last step are $$\mathrm{TRUE} \lor \mathrm{TRUE} \equiv \mathrm{TRUE}$$ but I don't think it will leave me to get the conjunction I'm looking for, right?

Since $$p\lor\neg p\equiv\mathrm{T}$$ and $$q\lor\neg q\equiv\mathrm{T}$$, you have \begin{align*} \mathrm{T}\lor\mathrm{T}&\equiv\mathrm{T}\\ &\equiv\mathrm{T}\land \mathrm{T}\\ &\equiv(p\lor\neg p)\land(q\lor\neg q). \end{align*}