Excuse me,how to do this type of question?

Using the laws of the algebra of sets show that

(a-b)∩(a∩b)'=a∩b'

Printable View

- May 25th 2010, 11:45 PMmayalgebraic laws of sets
Excuse me,how to do this type of question?

Using the laws of the algebra of sets show that

(a-b)∩(a∩b)'=a∩b' - May 26th 2010, 05:00 AMDeadstar
$\displaystyle (A \backslash B) \cap (A \cap B)'$

$\displaystyle = (A \cap B') \cap (A' \cup B')$

$\displaystyle = A \cap B' \cap (B' \cup A')$

$\displaystyle = A \cap B'$

Where we used the following rules...

$\displaystyle (A \backslash B) = (A \cap B')$

$\displaystyle (A \cap B)' = (A' \cup B')$

$\displaystyle B' \cap (B' \cup A') = B'$ - May 26th 2010, 05:01 AMWilmer
- May 26th 2010, 06:43 AMGrandad
Hello everyone

All that Deadstar says is correct, but I think the statement$\displaystyle B' \cap(B'\cup A') = B'$(although obvious) requires further proof - it's not one of the usual Set Laws (although some people do include it as an*Absorption Law*).

You prove it like this:$\displaystyle B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$Grandad$\displaystyle =B'\cup(\oslash \cap A')$

$\displaystyle =B'\cup\oslash$

$\displaystyle =B'$

- May 26th 2010, 06:45 AMmay
$\displaystyle (A \backslash B) = (A \cap B')$

$\displaystyle (A \cap B)' = (A' \cup B')$

$\displaystyle B' \cap (B' \cup A') = B'$[/quote]

how to know when to apply this rule? - May 26th 2010, 06:48 AMGrandad
- May 26th 2010, 06:56 AMmay
- May 26th 2010, 06:59 AMmay
- May 26th 2010, 07:05 AMGrandad
Hello mayThis uses the Distributive Law:

$\displaystyle (P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)$where I have replaced $\displaystyle P$ by $\displaystyle B'$, $\displaystyle Q$ by $\displaystyle \oslash$ and $\displaystyle R$ by $\displaystyle A'$.

Yes, I'm rather afraid that you do. As I said - you'll need lots of practice!

Grandad

PS ... and to answer your extra question, following your edit:

$\displaystyle \oslash \cap A' = \oslash$ uses what I have always known as an*Identity Law*, but others sometimes refer to as a*Domination Law*.

See, for example, here. - May 27th 2010, 08:11 AMmay
- May 27th 2010, 08:22 AMmay
- May 27th 2010, 01:31 PMGrandad