If $\{P \lor Q\} \land \{Q \implies R\}$ is true, does it follow that $\{P \lor R\}$ is also true? If $$\{P \lor Q\} \land \{Q \implies R\}$$ is true, does it follow that
$$\{P \lor R\}$$
is also true?
Here is my attempt, using a truth table:
$$
\begin{array}{ccc|c|c|c|c}
P & Q & R & P \lor Q & Q \implies R & \{P \lor Q\} \land \{Q \implies R\} & P \lor R \\
\hline
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 & 0 & 0 & 1 \\
1 & 0 & 1 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
\end{array}
$$
(Note that $1$ represents a truth value of True, while $0$ represents a truth value of False.)
Does this conclusively prove my claim?
Update
I intentionally did not include the extra column for
$$\{\{P \lor Q\} \land \{Q \implies R\}\} \implies \{P \lor R\}$$
anymore because I could just read off the truth value of the implication from the juxtaposed truth values of the premise and conclusion.
At any rate, I noticed from the truth table that it is almost a biconditional, save for the second and eighth rows.  Any idea on where the conjectured biconditional breaks down?  =)
 A: It almost proves your claim — it provides all the evidence you need. What you're trying to show is that the following formula is a tautology, true for all values of $P,Q,R$:
$$
((P \lor Q) \land (Q \!\implies\! R)) \implies (P \lor R).\tag{*}
$$
You can either add another column to your truth table for this formula, and confirm that every value is $1$, or you can reason as follows:
The formula (*) is a tautology iff for every row of the truth table, the value in the $((P \lor Q) \land (Q \!\implies\! R))$ column is $\le$ the value in the $(P \lor R)$ column. By inspection, that's the case.
There are less tedious ways to prove this. If $(P\lor R)$ is false, then both $P$ and $R$ have to be false; by hypothesis, $Q\!\implies\! R$ is true, so $Q$ has to be false for that to be true; but then $P\lor Q$ is false. So if the premises are true, then $P\lor R$ has to be true too.
A: What you need to do is to prove 
$$
((P∨Q)∧(Q⟹R))⟹(P∨R)\tag1
$$
is a tautology. You'd better do it by Boolean algebra rather than truth table. 
\begin{align}
(1)&\iff(\neg((P∨Q)∧(\neg Q∨R))∨(P∨R))
\\
&\iff((\neg(P∨Q)∨\neg(\neg Q∨R))∨(P∨R))
\\
&\iff((\neg P∧\neg Q)∨(Q∧\neg R))∨(P∨R))
\\
&\iff(\neg P∧\neg Q)∨((Q∨P∨R)∧(\neg R∨P∨R))
\\
&\iff(\neg P∧\neg Q)∨((Q∨P∨R)∧1)
\\
&\iff(\neg P∧\neg Q)∨(Q∨P∨R)
\\
&\iff(\neg P∨Q∨P∨R)∧(\neg Q∨Q∨P∨R)
\\
&\iff 1∧1
\\
&\iff 1
\end{align}
So $(1)$ is a tautology.
We also have following equivalent form
\begin{align}
(P∨R)&\iff((P∨R)∨1)
\\
&\iff((P∨R)∨(Q∧\neg Q))
\\
&\iff(P∨R∨Q)∧(P∨R∨\neg Q)
\\
&\iff(\neg(P∨R)⟹Q)∧(Q⟹(P∨R))
\end{align}
A: Yes, your truth table proves the claim.
We can also observe: $P \vee Q$ is equivalent to $\neg P \implies Q$.  With $Q \implies R$, that gives $\neg P \implies R$, which is equivalent to $P \vee R$.
ETA: Regarding your last question, in the first case, a true value of $P$ "conceals" $\neg (Q \implies R)$, and in the second case, a true value of $R$ "conceals" $\neg P$.
A: Note: $x \land y$ being true means both $x$ and $ y$ are true.  So let $x = P \lor Q$ and $ (P \lor Q) \land y $ being true means $(P \lor Q)$ is true no matter what the **** P, Q, y are.
