Help with proof of distributive law in propositional logic I'm having immense trouble understanding the proof of: 
$$(P \vee Q) \wedge (P \vee R) \vdash P \vee (Q \wedge R)$$
With only $(P \vee Q)$ and $(P \vee R)$ as premises, I understand that I have to assume $(P \vee Q) \wedge (P \vee R) \vdash P \vee (Q \wedge R)$ in a subproof but not what I have to do from there, that is: I don't entirely understand how $(Q \wedge R)$, or $(P)$ for that matter, is ever proven.  Any help would be much appreciated.    
 A: Since the OP does not know how to begin starting the proof, I will provide two proofs as examples using a Fitch-style proof checker that the OP can use to check that the proofs are correct and for further practice.
The first proof is a direct proof. Starting with the premise, $(P∨Q)∧(P∨R)$, I will derive the conclusion, $P∨(Q∧R)$:

On lines 2 and 3, I derive each of the conjuncts in line 1 using conjunction elimination (∧E).
Since each of these are disjuncts, that is, two statements connected by "or", in order to use them I have to consider each of the cases and derive the same result.  I start with the first statement on line 2. I consider the $P$ case on lines 4 to 5 using disjunction introduction (vI) to add to $P$ exactly what I need to reach the goal with this case. On lines 6 to 12 I consider the $Q$ case. This is harder. To derive this I have to consider the statement on line 3 with its two cases, $P$ and $R$.
For each case, I was able to derive the desired result and so the proof succeeded.

The next proof approaches the the problem by assuming the same premises, but also assuming that the conclusion is false.  I need to derive an absurdity (⊥) or a contradiction between two statements for this to succeed.

The first three lines are similar to the direct proof. On line 4, I assume the negation of the conclusion is true. On line 5 I use De Morgan's law on line 4. That gives me a statement with two conjuncts. I derive each conjunct on lines 6 and 7. On line 8, I use disjunctive syllogism (DS) to derive the disjunct I need from line 2. On line 9, I do the same for line 3. After using conjunction introduction (∧I) on line 10, this gives me a statement that contracts another statement, line 7. I can now derive the absurdity on line 11. On line 12, I use indirect proof (IP) to complete the proof.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf
A: Basically,  $(P\vee Q)\wedge (P\vee R)$ means $P$ or $Q$, and $P$ or $R$.


*

*In the case of $P$ then we clearly derive $P\vee(Q\wedge R)$.

*In the case of $Q$ then $P\vee R$ is also still true


*

*We've dealt with the case of $P$.

*In the case of $R$ (under condition of $Q$) then we derive $P\vee (Q\wedge R)$.



So all cases derive $P\vee(Q\wedge R)$.
