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

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

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

Where we used the following rules...

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

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

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

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 You prove it like this:

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

Grandad\(\displaystyle =B'\cup(\oslash \cap A')\)

\(\displaystyle =B'\cup\oslash\)

\(\displaystyle =B'\)

\(\displaystyle =B'\cup\oslash\)

\(\displaystyle =B'\)

Sorry - there's no quick answer to that - just lots of practice!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?

Grandad

which formula Grandad are using to change frHello 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 anAbsorption Law).

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

\(\displaystyle =B'\cup\oslash\)

\(\displaystyle =B'\)

[/SIZE]

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

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

\(\displaystyle =B'\cup\oslash\)

\(\displaystyle =B'\cup\oslash\)

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HOW to do this type of question?Must we memorise All the formula?Hello maySorry - there's no quick answer to that - just lots of practice!

Grandad

Hello may

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.

This uses the Distributive Law:which formula grandad are using to change fr

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

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

\(\displaystyle (P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)\)

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

Grandad

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

\(\displaystyle \oslash \cap A' = \oslash\) uses what I have always known as an

See, for example, here.

Last edited:

Oh,thanks Grandad at last i start to figure out how to do this type of question.thank you very much(Wink)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 anIdentity Law, but others sometimes refer to as aDomination Law.

See, for example, here.

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