Hi!

[kiki123]

Does anyone know how to prove this (A\B) U (B\A) = (A U B) \ (A ∩ B)

 

[tkhunny]

There is an important rule, here. Do SOMETHING.

What have you tried?

How about an identity or two...let's see...\(\displaystyle A/B = A - A\cap B\)

That might lead to something.

Let's see what you get.

 

[pka]

Does anyone know how to prove this (A\B) U (B\A) = (A U B) \ (A ∩ B)

Well \(\displaystyle A\setminus B\) means \(\displaystyle A\cap B^c\)
So what does \(\displaystyle A\setminus B \cup B\setminus A=~?\)

 

[soroban]

Hello, kiki123!

Definition: .\(\displaystyle A \backslash B \:=\:A \cap \overline{B}\)

\(\displaystyle \text{Prove: }\A\backslash B) \cup (B\backslash A) \:=\: (A \cup B) \backslash (A \cap B)\)

On the right side:

,. . \(\displaystyle \begin{array}{ccccccc}1. & (A \cup B) \backslash (A \cap B) && 1. & \text{Given} \\ \\ 2. & (A\cup B) \cap (\overline{A\cap B})&& 2. & \text{De{f}inition} \\ \\ 3. & (A\cup B) \cap (\overline{A} \cup \overline{B}) && 3. & \text{DeMorgan} \\ \\ 4. & (A\cap \overline{A}) \cup (A\cap \overline{B}) \cup (B \cap \overline{A}) \cup (B \cap \overline{B}) && 4. & \text{Distributive} \\ \\ 5. & \emptyset \cup (A \cap \overline{B}) \cup (B \cap \overline{A}) \cup \emptyset && 5. & S \cap \overline{S} \:=\:\emptyset \\ \\ 6. & (A\cap \overline{B}) \cup (B \cap \overline{A}) && 6. & S \cup \emptyset \:=\:S \\ \\ 7. & (A \backslash B) \cup (B \backslash A) && 7. & \text{De{f}inition} \end{array}\)