Prove $P \lor (P \land Q) \equiv P$ without using a truth table I don't know how to solve the question above without using a truth table. If anyone could help me, that would be great! The method used should be used something like LHS and RHS.
 A: You use $\phi\rightarrow \phi\lor\psi$ (disjunction introduction) and $\phi\land\psi\rightarrow\phi$ (conjunction elimination) together with the distsributive law and idempotence:
$$\begin{align}
P\lor(P\land Q)
&\leftrightarrow (P\lor P)\land(P\lor Q) & \text{Distributive law}\\
&\leftrightarrow P\land (P\lor Q)&\text{Idempotence}\\
&\rightarrow P &{\phi\land\psi\rightarrow\phi}\\
&\rightarrow P\lor (P\land Q) & \phi\rightarrow\phi\lor\psi
\end{align}$$
Then of course you use that $\phi\rightarrow\psi\rightarrow\phi$ means that $\phi\leftrightarrow\psi$.
A: You have 2 cases:


*

*$\mathcal{P}$ is false $\Leftrightarrow (\mathcal{P} \wedge\mathcal{Q})$ is false $\Leftrightarrow \mathcal{P}\vee (\mathcal{P} \wedge\mathcal{Q})$ is false.

*$\mathcal{P}$ is true $\Leftrightarrow \mathcal{P}\vee (\mathcal{P} \wedge\mathcal{Q})$ is true.
A: $$P\land (P \lor Q) \Leftrightarrow \text{ (Identity)}$$
$$(P \lor \bot) \land (P \lor Q) \Leftrightarrow \text{ (Distribution)}$$
$$P \lor (\bot \land Q) \Leftrightarrow \text{ (Annihilation)}$$
$$P \lor \bot \Leftrightarrow \text{ (Identity)}$$
$$P$$
