TITLE: Showing $P \land (P \lor Q) = P$ using only laws of propositions.
QUESTION [5 upvotes]: How can I show $P \land (P \lor Q) = P$ using only laws of propositions?
I see how $P \land (P \lor Q) = P$ simplifies to just $P$ using a truth table, but I can't see how to get it using propositional calculus. The distributive law just changes it to $P \lor (P \land Q)$, leaving me essentially in the same position.

REPLY [2 votes]: Here is one way to show this using natural deduction:

The equivalence will be derived using the biconditional introduction inference rule. I will consider each side of the biconditional as separate subproofs and then reference both at the end.
I consider $(P\land (P\lor Q))\to P$ on lines 1-2. I assume the antecedent $(P\land (P\lor Q))$ on line 1 and derive the consequent $P$ using conjunction elimination ($\land$E) on line 2. 
I consider the other direction $P\to (P\land (P\lor Q))$ on lines 3-5. I assume the antecedent $P$ on line 3 and derive the consequent $(P\land (P\lor Q))$ using disjunction introduction ($\lor$I) on line 4 and conjunction introduction ($\land$I) on line 5.
Line 6 uses biconditional introduction on the two subproofs mentioned above  referencing each subproof as a range of lines.

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