Let $A,B,C,D \subseteq X$ Show that $(A \setminus B ) \bigtriangleup (C \setminus D) \subseteq (A \bigtriangleup C) \cup (B \bigtriangleup D)$

My advances

$(A \setminus B ) \bigtriangleup (C \setminus D) $

$\rightarrow \left [ (A \setminus B ) \cap (C \cap D^{c} )^{c} \right ] \cup$ $\left [ (C \setminus D ) \cap ( A \cap B^{c} )^{c} \right ] $

$\rightarrow \left [ (A \cap B^{c}) \cap (C \cap D^{c} )^{c} \right ] \cup$ $\left [ (C \cap D ^{c}) \cap ( A \cap B^{c} )^{c} \right ] $

$\rightarrow \left [ (A \cap B^{c}) \cap (C^{c} \cup D) \right] \cup \left [ (C \cap D ^{c}) \cap ( A^{c} \cup B ) \right ] $

I can't conclude..

I Think, that is necessary find ...

$(A \bigtriangleup C ) \cup (B \bigtriangleup D) $

$=\left [ (A \setminus C) \cup (C \setminus A ) \right ] \cup \left [ (B \setminus D) \cup (D \setminus B ) \right ] $

$\rightarrow \left [ (A \cap C^{c}) \cup (C \cap A ^{c}) \right ] \cup \left [ (B \cap D ^{c}) \cup ( D \cap B^{c}) \right ] $

but I can't find their relationship..... I need help. Please!!!

  • $\begingroup$ Try using an element argument. Let x be an element of the left hand side and unwrap the definitions of your set operations. $\endgroup$ – Danny Lara Sep 29 '15 at 5:51

\begin{align} (A \setminus B ) \bigtriangleup (C \setminus D)&=((A \cap B^{c}) \cap (C \cap D^{c} )^{c})\cup ((C \cap D^{c}) \cap ( A \cap B^{c} )^{c}) \\ &=((A \cap B^{c}) \cap (C^{c} \cup D)) \cup ((C \cap D^{c}) \cap ( A^{c} \cup B )) \\ &=(A \cap B^{c} \cap C^{c}) \cup (A \cap B^{c} \cap D) \cup (C \cap D^{c} \cap A^{c}) \cup (C \cap D^{c} \cap B) \\ &\subset (A \cap C^{c}) \cup (B^{c} \cap D) \cup (C \cap A^{c}) \cup (D^{c} \cap B)\tag1 \\ &=(A \setminus C)\cup (D \setminus B)\cup (C \setminus A)\cup (B \setminus D) \\ &=(A \bigtriangleup C) \cup (B \bigtriangleup D) \end{align} $(1)$ We have $A \cap B^{c} \cap C^{c}\subset A \cap C^{c}, \: A \cap B^{c} \cap D\subset B^{c} \cap D, \:C \cap D^{c} \cap A^{c}\subset C \cap A^{c}, \:C \cap D^{c} \cap B\subset D^{c} \cap B$

  • $\begingroup$ Thanks a lot of for shared us your knowledge $\endgroup$ – El Chapo Sep 29 '15 at 16:45

Another way to do it is to take an arbitrary element $x$ in the set on the left and show that it belongs to the set on the right.

Consider the case $x \in A \setminus B, x \not\in C \setminus D$. Now $x \in C \cap D$ or $x \notin C$. In the first case, $x \in D \Delta B$ and in the second $x \in A \Delta C$. The case when $x \in C \setminus D$, $x \notin A \setminus B$ is symmetric; you can also perhaps convert this proof in algebraic terms.


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