Is the venn diagram for $\displaystyle (A - (A \cap B)) \cup (B - (A \cap B)) $, the same as $\displaystyle (A-B) \cup (B-A)$
where as $\displaystyle A \cap B \neq \varnothing$
Hello, Discrete!
Here's another proof . . .
Prove: .$\displaystyle [A - (A \cap B)] \cup [B - (A \cap B)] \;=\;(A-B) \cup (B-A)$
We need (among other definitions and properties):
. . [1] .$\displaystyle P - Q \;=\;P \cap\, \overline{Q}$ . (Def. of set subtraction)
. . [2] .$\displaystyle P \cap \overline P \:=\:f$
. . [3] .$\displaystyle P \cup f \:=\:P$
We have: .$\displaystyle [A - (A \cap B)] \cup [B - (A \cap B)]$
. . . . . $\displaystyle = \;\left[A \cap (\overline{A \cap B})\right] \cup \left[B \cap \left(\overline{A \cap B}\right)\right]$ . . . . . . . . . . . . . . . [1]
. . . . . $\displaystyle = \;\left[A \cap \left(\overline A \cup \overline B\right)\right] \cup \left[B \cap \left(\overline A \cup \overline B\right)\right] $ . . . . . . . . . . . DeMorgan's Law
. . . . . $\displaystyle = \;\left[\left(A \cap \overline A\right) \cup \left(A \cap \overline B\right)\right] \cup \left[\left(B \cap \overline A\right) \cup \left(B \cap \overline B\right)\right]$ . . Distributive Property
. . . . . $\displaystyle = \;\left[f \cup \left(A \cap \overline B\right)\right] \cup \left[\left(B \cap \overline A\right) \cup f\right]$ . . . . . . . . . . . . . . . [2]
. . . . . $\displaystyle = \;\left(A \cap \overline B\right) \cup \left(B \cap \overline A\right)$ . . . . . . . . . . . . . . . . . . . . . . [3]
. . . . . $\displaystyle = \;(A - B) \cup (B - A)$ . . . . . . . . . . . . . . . . . . . . . . [1]