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.


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?

  • $\begingroup$ It's certainly possible, though you need additional laws such as De Morgan's law, as well the fact that $(B^c)^c = B$. $\endgroup$ – B. Mehta Apr 9 '18 at 0:47
  • 1
    $\begingroup$ The approach using complements works fine. Please show us your work so we can help you find the mistake. $\endgroup$ – John Douma Apr 9 '18 at 0:50

$$\begin{align} A - (A - B) &= A - (A \cap B^c) \\ &= A \cap (A \cap B^c)^c \\ &= A \cap (A^c \cup (B^c)^c) \\ &= A \cap (A^c \cup B) \\ &= (A \cap A^c) \cup (A \cap B) \\ &= A \cap B \end{align}$$

Using $(B^c)^c = B$, De Morgan's law, and distribution.


Notice $A - B = A \cap B^c $. Hence,

$$ A - (A - B) = A \cap ( A \cap B^c )^c = A \cap (A^c \cup (B^c)^c)$$

Where we use DeMorgan's Law in second equality. Obviusly, $(B^c)^c = B$. Thus,

$$ A \cap (A^c \cup (B^c)^c) = A \cap (A^c \cup B ) = (A \cap A^c) \cup (A \cap B) = \varnothing \cup (A \cap B) = A \cap B $$


Using that property and De Morgan's law:

$$ A-(A-B)=A-(A \cap B^c) = A \cap (A \cap B^c)^c$$ $$ = A \cap (A^c \cup B) = (A \cap A^c) \cup (A \cap B) = \varnothing \cup (A \cap B) =A \cap B $$


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