Re: How would you prove...

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Originally Posted by

**Cairo** How would you prove that

(A/B) U B = A if and only if B C A ?

There is no one way to do any set theory proof.

Here you might note that $\displaystyle \left( {A\backslash B} \right) \equiv \left( {A \cap B^c } \right)$

Starting with $\displaystyle B\subseteq A$ what would $\displaystyle \left( {A \cap B^c } \right)\cup A=~?$

Re: How would you prove...

Quote:

Originally Posted by

**Cairo** How would you prove that

(A/B) U B = A if and only if B C A ?

It's the iff that is giving me problems. Is there a technique for this type of proof?

This statement is equivalent to the following statement :

$\displaystyle (p\land \lnot q) \lor q \Leftrightarrow p$ iff $\displaystyle p \lor q \Leftrightarrow p$

Re: How would you prove...

Quote:

Originally Posted by

**princeps** This statement is equivalent to the following statement :

$\displaystyle (p\land \lnot q) \lor q \Leftrightarrow p$ iff $\displaystyle p \lor q \Leftrightarrow p$

Thanks, but this looks even worse to me!

Re: How would you prove...

Would this not be A, Plato?

I was thinking of relabelling (B C A) as D and ((A/B) U B = A) as E and then trying to argue D implies E and also that ~D implies ~E, but not sure if this would work or even where to start.

Re: How would you prove...

Quote:

Originally Posted by

**Cairo** Would this not be A, Plato?

That is correct. So you have proved it one way.

Now suppose that $\displaystyle (A\cap B^c)\cup B=A$ now show that $\displaystyle B\subseteq A.$

Re: How would you prove...

Quote:

Originally Posted by

**Plato** That is correct. So you have proved it one way.

Now suppose that $\displaystyle (A\cap B^c)\cup B=A$ now show that $\displaystyle B\subseteq A.$

Using the distributive law I get

(A U B) and (B^c U B) = .......

But not sure how to continue this argument.

Will have a think about it.

Re: How would you prove...

Quote:

Originally Posted by

**Plato** There is no one way to do any set theory proof.

Here you might note that $\displaystyle \left( {A\backslash B} \right) \equiv \left( {A \cap B^c } \right)$

Starting with $\displaystyle B\subseteq A$ what would $\displaystyle \left( {A \cap B^c } \right)\cup A=~?$

Should not $\displaystyle \left( {A \cap B^c } \right)\cup A$ be $\displaystyle \left( {A \cap B^c } \right)\cup B$

And then be equal to: $\displaystyle (B\cup A)\cap(B\cup B^c)$ ?