I'm studying for a mathematics class and have been struggling with the following proof
$$A-(A-B) = A ∩ B$$
I know we have to use the following rule $A-B = A ∩ B^c$ which is set $B$'s complement known as the set difference law.
Do I have everything necessary to solve this prove? Is this proof possible because using the complements won't seem to result in the $A ∩ B$ that is required.
Edit-
Thank you guys I gained some confidence and done the proof this is what I got
A-B = A ∩ B^(c) – Set Difference Law
A- (A ∩ B^(c)) = A ∩ B
A- (A ∩ B^(c)) = (A ∩ (A ∩ B^(c)) ^c) – Set Difference Law
(A ∩ (A ∩ B^(c)) ^c) = A ∩ B – double Complement Law
(A ∩ (A ∩ B)) = A ∩ B
A ∩ A = A – Idempotent Laws (Can I do this?)
A ∩ B = A ∩ B
I think this proof is close to correct. I was confused with the last part but it seems to check out.
Edit 2- Nevermind Looks like my proof was all wrong. Is there any way to redeem my current proof?
Best Answer
$$\begin{align} A - (A - B) &= A - (A \cap B^c) \\ &= A \cap (A \cap B^c)^c \\ &= A \cap (A^c \cup (B^c)^c) \\ &= A \cap (A^c \cup B) \\ &= (A \cap A^c) \cup (A \cap B) \\ &= A \cap B \end{align}$$
Using $(B^c)^c = B$, De Morgan's law, and distribution.