# Math Help - algebraic laws of sets

1. ## algebraic 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'

2. $(A \backslash B) \cap (A \cap B)'$

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

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

$= A \cap B'$

Where we used the following rules...

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

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

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

3. Hello everyone

All that Deadstar says is correct, but I think the statement
$B' \cap(B'\cup A') = B'$
(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:
$B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$
$=B'\cup(\oslash \cap A')$

$=B'\cup\oslash$

$=B'$

4. $(A \backslash B) = (A \cap B')$

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

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

how to know when to apply this rule?

5. Hello may
Originally Posted by may
Where we used the following rules...

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

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

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

how to know when to apply this rule?
Sorry - there's no quick answer to that - just lots of practice!

6. Originally Posted by Grandad
Hello everyone

All that Deadstar says is correct, but I think the statement
$B' \cap(B'\cup A') = B'$
(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:
$B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$
$=B'\cup(\oslash \cap A')$

$=B'\cup\oslash$

$=B'$

which formula Grandad are using to change fr

[/SIZE][INDENT] $B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$
[INDENT] $=B'\cup(\oslash \cap A')$

$=B'\cup\oslash$
[SIZE=3]

7. Originally Posted by Grandad
Hello maySorry - there's no quick answer to that - just lots of practice!

HOW to do this type of question?Must we memorise All the formula?

8. Hello may
Originally Posted by may
which formula grandad are using to change fr
$B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$

$=B'\cup(\oslash \cap A')$
This uses the Distributive Law:
$(P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)$
where I have replaced $P$ by $B'$, $Q$ by $\oslash$ and $R$ by $A'$.
Originally Posted by may
HOW to do this type of question?Must we memorise All the formula?
Yes, I'm rather afraid that you do. As I said - you'll need lots of practice!

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

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

See, for example, here.

9. Originally Posted by Grandad
Hello mayThis uses the Distributive Law:
$(P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)$
where I have replaced $P$ by $B'$, $Q$ by $\oslash$ and $R$ by $A'$.

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

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

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

See, for example, here.
Oh,thanks Grandad at last i start to figure out how to do this type of question.thank you very much

10. Originally Posted by Grandad
Hello mayThis uses the Distributive Law:
$(P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)$
where I have replaced $P$ by $B'$, $Q$ by $\oslash$ and $R$ by $A'$.

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

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

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

See, for example, here.
Oh,thanks Grandad at last i start to figure out how to do this type of question.thank you very much

erm i want to ask another question..same topic do i need to put the question in new thread or at here?

11. Hello may
Originally Posted by may
Oh,thanks Grandad at last i start to figure out how to do this type of question.thank you very much

erm i want to ask another question..same topic do i need to put the question in new thread or at here?
Start a new thread.