Use set identities to prove....

[hank]

Use set identities to prove:

"For all sets A, B, and C that are subsets of the universe U, if A intersect B = A intersect C and ~A intersect B = ~A intersect C, then B = C".

Ok, my professors says we can only use set identities for this proof.

My problem is that there are no similar homework problems, and there are no examples in the book that are similar.
As a result, I'm really stuck.

We are permitted to use a couple theorems that lay out some equivalents, i.e. A intersect A = A.

I'm not sure where to even start. I think I could do the proof if I wasn't constrained in such a way.

Can someone give me an example from somewhere?
Or perhaps give a better explanation of what she means?

The only thing I can think of is to start something like...

~A intersect B = B - A

But then, I don't know what else to do.
I'm not sure how to even structure the proof.

Edit: Here's the best that I could come up with:

Assume A intersect B = A intersect C and A complement intersect B = A complement intersect C.
By identity, A complement intersect B = B - A and A complement intersect C = C - A.
So, B - A = C - A.
Thus, A is not a subset of B and A is not a subset of C.
Since A intersect B = A intersect C and A is not a subset of B and A is not a subset of C, then B = C.

I'm not asking if this answer is right or wrong, but what I am asking is if this is a suitable type of answer for the question?

 

[pka]

You are given that \(\displaystyle A \cap B = A \cap C\,\& \,A^c \cap B = A^c \cap C\)
Now suppose B contains some element not in C: \(\displaystyle \left( {\exists x \in B\backslash C} \right)\).
There are two possiblites: \(\displaystyle \left( {x \in A} \right) \vee \left( {x \in A^c } \right)\).
\(\displaystyle \left( {x \in A} \right) \Rightarrow \quad x \in A \cap B = A \cap C \Rightarrow \quad x \in C\)
That is a contradiction. This is also a contradiction:
\(\displaystyle \left( {x \in A^c } \right) \Rightarrow \quad x \in A^c \cap B = A^c \cap C \Rightarrow \quad x \in C\)
Thus \(\displaystyle B \subseteq C\) likewise we prove \(\displaystyle C\subseteq B\).

 

[hank]

Thank you for the reply.

However, I'm only allowed to use set identities.
I am not allowed to do the proof using definitions.

 

[pka]

\(\displaystyle B=(B\cap A)\cup(B\cap A^c)=(C\cap A)\cup(C\cap A^c )=C\)

 

[hank]

Ahhhhh....

Thanks.

I would never have thought to go about it that way.