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Lets define $A\circ B$ as the symmetrical difference with $$A\circ B = (A\setminus B)\cup(B\setminus A) = (A\cup B)\setminus (A\cap B)$$

Show that $(A\circ B)\cup C = (A\cup B)\circ (B\cup C)$ without using the distributive law of sets and without the laws of the difference between two sets. Also not using the laws of DeMorgan.

I do not find any way without using one of them

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  • $\begingroup$ Please format your question appropriately. $\endgroup$ – user8960 Oct 24 '16 at 21:57
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    $\begingroup$ If you want to demonstrate equality, establish subset in both directions by taking an element of the left-hand side and showing that it is also an element of the right-hand side, and vise versa. $\endgroup$ – Ispil Oct 24 '16 at 22:00
  • $\begingroup$ @lspil i tried that but i did not find a way to show that x is in the left side and then also in the right $\endgroup$ – Arjihad Oct 24 '16 at 22:02
  • $\begingroup$ Why is it tagged as linear algebra question? $\endgroup$ – Arthur Oct 24 '16 at 22:17
  • $\begingroup$ Start like this: Suppose $x \in (A \circ B) \cup C$. Then what do you know? First, you know that either $x \in A\circ B$ or $x \in C$. What else? Just take steps like this until you conclude that $x \in (A\cup B)\circ (B\cup C)$ $\endgroup$ – Alex G. Oct 24 '16 at 22:28

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