- #1

SamRoss

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- #1

SamRoss

Gold Member

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- #2

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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*}

- #3

Stephen Tashi

Science Advisor

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

- #4

FactChecker

Science Advisor

Gold Member

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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.

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