# Prove a set Identity

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1. Aug 8, 2015

### 22990atinesh

1. The problem statement, all variables and given/known data
Prove that $A\cap(B\Delta C)=(A\cap B)\Delta(A\cap C)$

2. Relevant equations

3. The attempt at a solution

L.H.S.=$A\cap(B\Delta C)$
=$A\cap[(B - C) \cup (C - B)]$
=$A\cap[(B \cap \bar{C}) \cup (C \cap \bar{B})]$
=$[A\cap (B \cap \bar{C})] \cup [A\cap (C \cap \bar{B})]$
=$[(A\cap B) \cap \bar{C}] \cup [(A\cap C) \cap \bar{B}]$

R.H.S.=$(A \cap B) \Delta (A \cap C)$
=$[(A \cap B) - (A \cap C)] \cup [(A \cap C) - (A \cap B)]$
=$[(A\cap B) \cap \bar{(A\cap C)} ] \cup [(A\cap C) \cap \bar{(A\cap B)}]$
=$[(A\cap B) \cap (\bar{A} \cup \bar{C})] \cup [(A\cap C) \cap (\bar{A} \cup \bar{B})]$

I tried to make L.H.S = R.H.S. But with above results, It's not possible. Can anyone tell me what I've assumed wrong.

2. Aug 8, 2015

### HallsofIvy

Staff Emeritus
Are you required to use that method? The most basic way to prove "X= Y" is to prove both "$X\subseteq Y$" and "$Y\subseteq X$"".
And the most basic way to prove "$X\subseteq Y$" is to start "if x in in X" and use the definitions and properties of X and Y to conclude "therefore x is in Y".

Here we want to prove $A\cap\left(B\Delta C\right)= \left(A\cap B\right)\Delta\left(A\cap C\right)$ so we first prove
$A\cap\left(B\Delta C\right)\subseteq \left(A\cap B\right)\Delta\left(A\cap C\right)$

To do that:
if $x\in A\cap\left(B\Delta C\right)$ then x is in A and x is in B or C but not both. So look at two cases

1) x is in B but not C. Then x is in $A\cap B$ but not $A\cap C$. Therefore x is in $\left(A\cap B\right)\Delta\left(A\cap C\right)$.

2) x is in C but not in B. Then x is in $A\cap C$ but not $A\cap B$. Therefore x is in $\left(A\cap B\right)\Delta\left(A\cap C\right)$.

Now show that $\left(A\cap B\right)\Delta\left(A\cap C\right)\subseteq A\cap\left(B\Delta C\right)$ the same way:
if $x \in \left(A\cap B\right)\Delta\left(A\cap C\right)$ then ...

3. Aug 8, 2015

### 22990atinesh

I understand your approach. But I want to prove it through Set identities.