sets prove

**thereddevils** · May 29th 2009, 04:39 AM

Show that for any sets A and B ,

(A-B)U(B-A)=(AUB)-(AnB)

thanks a lot ..

**Soroban** · May 29th 2009, 07:03 AM

Hello, thereddevils!

Show that for any sets $A$ and $B\!:\;\;<br /> (A-B) \cup (B-A) \:=\:(A \cup B)-(A \cap B)$

I'll establish some rules for reference:

. . $[1]\;\;P - Q \:=\:P \cap \overline{Q}$ . (definition of set subtraction)

. . $[2]\;P \cup \overline P \:=\:U$

. . $[3]\;\;P \cap U \:=\:P$

$\begin{array}{ccc}(A - B) \cup (B - A) & &\text{Given} \\ \\ (A \cap \overline B) \cup (B \cap \overline A) & & [1] \\ \\ \bigg[A \cup(B\cap \overline A)] \cap \bigg[\overline B \cup (B \cap \overline A)\bigg] & & \text{distr.} \\ \\ \bigg[(A \cup B) \cap (A \cup \overline A)\bigg] \cap \bigg[(\overline B \cup B) \cap (\overline B \cup \overline A)\bigg] & & \text{distr.} \\ \end{array}$

. . . $\begin{array}{cccc} \bigg[(A \cup B) \cap U\bigg] \cap \bigg[U \cap (\overline B \cup \overline A)\bigg]\qquad && & [2] \\ \\ (A \cup B) \cap (\overline A \cup \overline B) & & & [3] \\ \\ (A \cup B) \cap (\overline{A \cap B}) & & & \text{DeMorgan} \\ \\ (A \cup B) - (A \cap B) & & & [1] \end{array}$

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