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.

  • $\begingroup$ 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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  • $\begingroup$ 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

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