Proof of $(A \setminus B)\setminus C = (A \setminus C) \setminus (B \setminus C)$

Let $$A,B,C$$ be sets, show that then $$(A \setminus B)\setminus C = (A \setminus C) \setminus (B \setminus C)$$.

Using laws of associativity, distribution, De Morgan, I need to show this identity holds.

I think my final destination is:

$$(A \cap B^\mathsf{c}) \cap (B \cap C^\mathsf{c})$$

I am following these steps: \begin{align} (A \setminus B) \setminus C &=\\ (A \cap B^\mathsf{c}) \cap C^\mathsf{c} &= \end{align}

• Try to write each of the $-$ signs in set notation. For two sets, $X - Y$ is what as sets? That is, it contains all elements that are in $X$ but not in $Y$, but how can you write this with union/intersection? If you do this, you can simplify what you get with De Morgan maybe. – астон вілла олоф мэллбэрг Feb 26 at 13:09
• DeMorgan: $(B\cap C)^C=B^C\cup C^C$ – J. W. Tanner Feb 26 at 13:13
• I don't think you want $B^C \cap B = \phi$ – J. W. Tanner Feb 26 at 13:14
• The LHS is correct : $A \cap B^\mathsf{c} \cap C^\mathsf{c}$. – Mauro ALLEGRANZA Feb 26 at 13:20
• The RHS is $(A \cap C^\mathsf{c}) \cap (B \cap C^\mathsf{c})^\mathsf{c}$ and using De Morgan : $(A \cap C^\mathsf{c}) \cap (B^\mathsf{c} \cup C)$ – Mauro ALLEGRANZA Feb 26 at 13:21

\begin{align} (A - C) - (B - C) &= (A \cap C^\mathsf{c}) \cap (B \cap C^\mathsf{c})^\mathsf{c}\\ &=(A \cap C^\mathsf{c}) \cap (B^\mathsf{c} \cup C)\\ &=[(A \cap C^\mathsf{c}) \cap B^\mathsf{c}] \cup [(A \cap C^\mathsf{c}) \cap C]\\ &=[A \cap C^\mathsf{c} \cap B^\mathsf{c}] \cup [\emptyset]\\ &=A \cap C^\mathsf{c} \cap B^\mathsf{c}\\ &=[A \cap B^\mathsf{c}] \cap C^\mathsf{c}\\ &=[A - B] - C \end{align} I have implicitly used the associative property of union and intersection. Also if you just want the validity of the statement, you can just use Venn's diagram without doing this long proof.
• On the first (and second) line, don't you mean $(B\cap C^C)^C$? – J. W. Tanner Feb 26 at 13:51
• Also, why didn't you say $[A \cap B^C] \cap C^C$ on the penultimate line? – J. W. Tanner Feb 26 at 13:54