High School How does r∪(-p∩q∩-r) simplify to r∪(-p∩q) ?

[SamRoss]

How does r∪(-p∩q∩-r) simplify to r∪(-p∩q) ? The second expression is just the first with the "-r" gone at the end. I'm not seeing how to get from the first expression to the second using any of the basic laws like distribution, de morgan, tautology, etc. What am I missing?

 

[andrewkirk]

Let $U$ be the set of which all these sets r,p,q are subsets, so that $-r=U-r$ and $U=r\cup -r$.
Then we have:
\begin{align*}
r\cup(-p\cap q\cap -r)
&= r\cup((-p\cap q)\cap -r) \quad\textrm{[associative law]}
\\&=
(r\cup(-p\cap q)) \cap( r\cup-r)
\quad\textrm{[distributive law]}
\\&=
(r\cup(-p\cap q)) \cap U
\quad\textrm{[see preamble above]}
\\&=
r\cup(-p\cap q)
\quad\textrm{[since $U\cap A=A$ for any $A\subseteq A$]}
\end{align*}

 

[Stephen Tashi]

After you do distribution, you can represent $(r \cup \neg r)$ as "$T$" since its always true. Then $(A) \cap T$ is equivalent to $(A)$.

Your text materials may use a different technique than the notation "T". They should have some rule equivalent to the effect of using "T".

 

[FactChecker]

The second expression will include elements of $(-p \cap q)$, whether they are in r or not. The first expression will only add elements of $(-p \cap q)$ that are not in r, but the others were in r already. So there is no difference.

To mathematically show that, you can break up $(-p \cap q) = (-p \cap q \cap r) \cup (-p \cap q \cap -r)$.
So $$ r \cup (-p \cap q) = r \cup ((-p \cap q \cap r) \cup (-p \cap q \cap -r)) $$
$$ = (r \cup (-p \cap q \cap r)) \cup (r \cup (-p \cap q \cap -r)) $$
$$ = r \cup (r \cup (-p \cap q \cap -r)) $$
$$ = r \cup (-p \cap q \cap -r) $$
There may be more direct ways to do this, but I don't see any now.