# algebraic laws of sets

#### may

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'

#### Deadstar

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

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

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

$$\displaystyle = A \cap B'$$

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

may

#### Grandad

MHF Hall of Honor
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:
$$\displaystyle B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$$
$$\displaystyle =B'\cup(\oslash \cap A')$$

$$\displaystyle =B'\cup\oslash$$

$$\displaystyle =B'$$

Grandad

may

#### may

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

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

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

how to know when to apply this rule?

#### Grandad

MHF Hall of Honor
Hello may
Where we used the following rules...

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

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

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

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

Grandad

may

#### may

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

$$\displaystyle =B'\cup\oslash$$

$$\displaystyle =B'$$

Grandad
which formula Grandad are using to change fr

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

$$\displaystyle =B'\cup\oslash$$

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#### may

Hello maySorry - there's no quick answer to that - just lots of practice!

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

#### Grandad

MHF Hall of Honor
Hello may
which formula grandad are using to change fr
$$\displaystyle B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$$

$$\displaystyle =B'\cup(\oslash \cap A')$$
This 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 $$\displaystyle A'$$.
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!

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.

Last edited:
may

#### may

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.
Oh,thanks Grandad at last i start to figure out how to do this type of question.thank you very much(Wink)

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