2
$\begingroup$

I am trying to write the statement above using logic. I managed to break up the first difference sign using the definition of set difference and get:

Suppose $$x\in (A-C) \land x\notin(B-C),$$

but I have no idea how to break it further.

$\endgroup$
  • $\begingroup$ Use U - V = U $\cap V^c$ and DeMorgan rules. $\endgroup$ – William Elliot Feb 10 '18 at 19:48
  • $\begingroup$ sorry my bad, I copied the initial equation wrong. It's (A-C)-(B-C) $\endgroup$ – mdrjjn Feb 10 '18 at 19:52
3
$\begingroup$

$x \in (A-C)-(B-C)\tag{(1) given}$

$\iff x \in A \land x \notin C \land \lnot(x \in B \land x \notin C)\tag{(2) def. set-difference}$

$\iff x\in A \land x \notin C \land (x \notin B \lor x \in C)\tag{(3) DeMorgan's}$

$\iff (x \in A \land x \notin C \land x \notin B) \lor\; \underbrace{\color{grey} { ( x \in A \land x \notin C \land x \in C)}}_{\large\varnothing} \tag{(4) distribution}$

$$ \iff x \in A \land x \notin C \land x \notin B\tag{(5)$\color{grey}{\text{Contradiction in} (4)}$}$$

$$ \iff x\in A \land x \notin B \land x\notin C\tag{(5) commutativity}$$

$\iff x\in A \land \lnot (x \in B \lor x \in C)\tag{(6) DeMorgan's }$

$\iff x \in A \land x \notin (B\cup C)\tag{(7) def. Set union}$

$$\iff x \in A - (B\cup C)\tag{(8), def. set difference}$$

Hence $$(A-C)- (B-C) = A-(B\cup C)$$

| cite | improve this answer | |
$\endgroup$
5
$\begingroup$

@amWhy has an excellent answer to this, but as an alternative you can also work directly with the set operators:

$$(A - C) - (B - C) = $$

$$(A \cap C^C) \cap (B \cap C^C)^C=$$

$$A \cap C^C \cap (B^C \cup C)=$$

$$A \cap C^C \cap B^C=$$

$$A \cap (C \cup B)^C=$$

$$A - (C \cup B)$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! Just starting out with this so this one looks slightly more complicated but definitely something I should strive to learn $\endgroup$ – mdrjjn Feb 10 '18 at 20:26
  • $\begingroup$ @mdrjjn Yeah, it's good to know how to do these ... the good news is that all the operations are exactly isomorph to what you do with the logical operations. :) Also, if amWhy's answer is satisfactory to you, you can accept it by checking on the check mark next to it. $\endgroup$ – Bram28 Feb 10 '18 at 20:52
  • $\begingroup$ Nice work. It's nice to have two approaches to suggest, and you're right, re: the correlation/isomorphism between dealing via logic/set operations! +1 $\endgroup$ – amWhy Feb 10 '18 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.