Is the following propositional logic valid? $$P ∧ (¬P∨Q) ∧ (¬Q∨R) → R$$
Truth table method seems to be hectic. Is there any other simple approach to check ?
 A: If suffices to check the following: if $R$ is false, then $P ∧ (¬P∨Q) ∧ (¬Q∨R)$ is false. This is because $x \rightarrow \text{True}$ is always valid.
Note that by the distributive property, $P ∧ (¬P∨Q) = (P∧¬P) ∨ (P∧Q) = P∧Q$. Also since $R$ is false, $¬Q∨R = ¬Q$. 
$$ P ∧ (¬P∨Q) ∧ (¬Q∨R) = (P∧Q) ∧ ¬Q = P∧(Q ∧ ¬Q)$$
Clearly, $Q ∧ ¬Q$ is false. This completes the proof. 
A: You have for the left hand side to be true that $P$ so $\neg P\lor Q$ is the same as $Q$ which is also required to be true so $\neg Q \lor R$ is $R$ so $R$ is also required to be true. This boils down to prove that:
$$P\land Q\land R\rightarrow R$$
Another way to see it is that $\phi\rightarrow\psi$ is the same as $\neg\phi\lor\psi$. Which makes the statement to be shown is the same as:
$$P\land (P\rightarrow Q)\land (Q\rightarrow R) \rightarrow R$$
which should be obvious.
A: Recall that $A\rightarrow B$ is equivalent to $\neg (A\wedge \neg B)$. Therefore, it suffices to check that whenever the left hand side of this implication is true, the right hand side must also be true. Since the left hand side is made up of the conjunction of three propositions, it is true only when all three conjuncts are true. For the first conjunct to be true, $P$ is true. For the second conjunct to be true, since $\neg P$ is necessarily false, $Q$ must be true. Finally, since $\neg Q$ must be false, $R$ must be true, which completes the proof.
A: Yes, this implication does hold. 
One way to see this is because the left-hand side is only true when all of $P, Q, $ and $R$ are true. In all other cases, the left-hand side is false, and the implication follows vacuously.
Concretely, if $P$ is false, then $P \land x$ is false for any statement $x$. 
Otherwise, suppose $P$ is true. Then if $Q$ is false, $\neg P \lor Q$ is false, rendering the entire left-hand side false (regardless of what $R$ is).
Finally, suppose $P$ and $Q$ are true. Then if $R$ is false, $\neg Q \lor R$ is false.
The only remaining case is that $P,Q$, and $R$ are all true, and you can check that in this case both sides of the implication are true. This rules out a statement of the form $F \implies T$ occurring in the truth table, and so the implication holds!
A: $$(P \land (\neg P \lor Q) \land (\neg Q \land R)) \rightarrow R \Leftrightarrow \text{ (Reduction)}$$
$$(P \land Q \land (\neg Q \land R)) \rightarrow R \Leftrightarrow \text{ (Reduction)}$$
$$(P \land Q \land R) \rightarrow R \Leftrightarrow \text{ (Implication)}$$
$$\neg (P \land Q \land R) \lor R \Leftrightarrow \text{ (DeMorgan)}$$
$$\neg P \lor \neg Q \lor \neg R \lor R \Leftrightarrow \text{ (Complement)}$$
$$\neg P \lor \neg Q \lor \top \Leftrightarrow \text{ (Annihilation)}$$
$$\top$$
So yeah, this is a valid statement (this was obvious after 2 steps already, but  I am not familiar with a rule $(X \land Y) \rightarrow X \Leftrightarrow \top$ ... there really should be!)
