Set Theory Theorems: Solving for A in A ∩ B = C ∩ B and A ∩ B' = C ∩ B

[mbcsantin]

I need help on how to get started with this question:
Im stocked and i just don't have a clue on how to figure this out.

Prove:
If A intersect B = C intersect B and A intersect B' = C intersect B' then A = C

 

[mathman]

To save writing * = intersec and + = union.

A = A*B + A*B' = C*B + C*B' = C

A = A*B + A*B' results from 2 things, distributive law for sets [A*(B+B') = A*B + A*B'] and the fact that B+B' is the entire space.

 

[mbcsantin]

To save writing * = intersec and + = union.

A = A*B + A*B' = C*B + C*B' = C

A = A*B + A*B' results from 2 things, distributive law for sets [A*(B+B') = A*B + A*B'] and the fact that B+B' is the entire space.

Thank you but
i'm a little bit confused now..
Let me just translate what you wrote,

So,
A = A*B + A*B' = C*B + C*B' = C

becomes

A = (AnB) U (AnB') = (CnB) U (CnB') = C the symbol "n" for intersect

A = AnB U AnB' results from 2 things, distributive law for sets [An(BUB') = AnB U AnB']

I just don't quite get it. Could you please justify it. Like step by step if possible?

Let me just re-write the question:
If A n B = C n B and A n B' = C n B' then A = C

 

[mathwonk]

they share the same points inside B and also share the same points outside B. E.g. if you and your roommate have the same girlfriends both in class and outside class, then you have all the same girlfriends.

 

[mbcsantin]

they share the same points inside B and also share the same points outside B. E.g. if you and your roommate have the same girlfriends both in class and outside class, then you have all the same girlfriends.

I understand the example you mentioned above but i still don't get this:

A = (AnB) U (AnB') = (CnB) U (CnB') = C

i don't get how you figure that out!
still doesn't make sense to me

 

[mathman]

In words A intersect B means all point in A and in B, while A intersect B' means all points in A and not in B. Put them together and you get all points in A.