I can't figure out whether my answer is correct or not. Consider part c of the following question -

I understand why $p \land (q \lor r)$ is correct. Am I correct in saying $(p \land q) \lor r$ is a correct answer as well?
 A: It is correct in the sense that it is true for the same $x$ that the inequality is, but it is not generally true that $(p \wedge q) \vee r$ is equivalent to $p \wedge (q \vee r)$  Having $r$ true and $p$ false makes the first true but not the second.  With the given $p,r$ this is not possible.  The book answer is the natural way to write it because the parenthesized part corresponds to the $\leq$ sign in the question.
Whether it is correct depends on what the teacher is looking for.
A: It happens that for these particular propositions, $p \lor r$ is equivalent to $p.$
It follows that $p \land (q \lor r)$ is equivalent to $(p \land q) \lor r,$
although you can also find this directly (as I suppose you did).
A: *

*A way to answer the question is to make the truth table can help you answer that question. You can construct the truth table for the formulation $p\wedge (q\vee r)$ and then construct the truth table for the formulation $(p\wedge q)\vee r$ and see if the table are the same (i,e., logically equivalent).

*On other hand, the distributive laws says $p\wedge (q\vee r)\equiv(p\wedge q)\vee (p\wedge r)$ and $p\vee(q\wedge r)\equiv (p\vee q)\wedge (p\vee r)$ and the associative laws says $(p\wedge q)\wedge r\equiv p\wedge (q\wedge r)$ and $p\vee (q\vee r)\equiv (p\vee q)\vee r$.

A: $P=\{x| 0<x\}$, $Q=\{x| x<3\}$ and $R=\{x| x=3\}$, and $p$ is the logic statement "$x\in P$", $q$ is the logic statement "$x\in Q$", $r$ is the logic statement "$x\in R$".
a) "$x\in Q\cup R$" is the logic statement $q\lor r$.
b) "$x\in P\cap Q$" is the logic statement $p\land q$.
c) "$x\in P\cap(Q\cup R)$" is the logic statement $p\land (q\lor r)$.
Since $P\cap(Q\cup R)=(P\cap Q)\cup (P\cap R)=(P\cap Q)\cup R$, here, $p\land (q\lor r)\equiv (p\land q)\lor r$. So, you are correct.
A: Just to be thorough the eight truth values of $p\land (q \lor r)$ and $(p \land q)\lor r$ are
$p=T;q=T;r=T$ then $p\land (q \lor r)=T$ and $(p \land q)\lor r=T$.  This happens where $0<x$ and $x<3$ and $x=3$.  In other words, this never happens.
$p=T;q=T;r=F$ then $p\land (q \lor r)=T$ and $(p \land q)\lor r=T$. This happens when $0 < x < 3$.
$p=T;q=F;r=T$ then $p\land (q \lor r)=T$ and $(p \land q)\lor r=T$. This happens when $0< x$ but $x\ge 3$ and $x=3$ or in other words: when $x = 3$.
$p=T;q=F;r=F$ then $p\land (q \lor r)=F$ and $(p \land q)\lor r=F$. This happens when $0< x$ but $x\ge 3$ and $x\ne 3$ or in other words: when $x> 3 $.
$p=F;q=T;r=T$ then $p\land (q \lor r)=F$ and $(p \land q)\lor r=T$.  !!THEY DISAGREE!!. This happens when $0\ge x$ but $x< 3$ and $x= 3$ or in other words, this NEVER happens! (so it doesn't matter that the disagree.)
$p=F; q=T;r=F$ then $p\land (q\lor r)=F$ and $(p\land q)\lor r=F$.  This happens when $0 \ge x$ and $x<3$ are $x\ne 3$ or in other words when $x\le 0$.
$p=F; q=F; r=T$ then $p\land (q\lor r)=F$ and $(p\land q)\lor r = T$.  !!THEY DISAGREE!!.  But this only happens when $x\le 0$ and $x \ge 3$ and $x = 3$.  This never happens.
$p=F; q=F; r=T$ then $p\land (q\lor r)=F$ and $(p\land q)\lor r =F$.  This happens when $x \le 0$ and $x \ge 0$ and $x\ne 3$.  This never happens.
....
So of the $8$ hypothetical truth values only $4$ of them are actually possible.  Your two statements are in complete agreement for those for cases.
