proof of $A\cap(B \triangle C)=(A\cap B)\triangle (A\cap C)$ using DeMorgan's laws I could prove this by using Venn Diagrams, but I couldn't use DeMorgan's laws to prove it.
I also used this method:


  
    
      
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How can I prove it by using DeMorgan's laws?
Any hints would be appreciated. Many thanks!
 A: Note that
\begin{align*}
(A\cap B)\triangle (A\cap C)
&= ((A\cap B)\cup(A\cap C))\setminus((A\cap B)\cap(A\cap C)) \\
&= (A\cap(B\cup C)) \setminus (A\cap B\cap C) \\
&= (A\cap(B\cup C)) \cap (A\cap B\cap C)^c \\
&\stackrel{\cdot}{=} (A\cap(B\cup C)) \cap(A^c\cup B^c\cup C^c) \\
&= (A\cap(B\cup C)\cap A^c)\cup(A\cap(B\cup C)\cap B^c)\cup(A\cap (B\cup C)\cap C^c) \\
&= (A\cap(B\cup C)\cap B^c)\cup(A\cap (B\cup C)\cap C^c) \\
&= A\cap(B\cup C)\cap(B^c\cup C^c) \\
&\stackrel{\cdot}{=} A\cap(B\cup C)\cap(B \cap C)^c \\
&= A\cap ((B\cup C)\setminus (B\cap C)) \\
&= A\cap (B\triangle C).
\end{align*}
The equalities with dots over them are applications of De Morgan's laws. It is much easier to read, and not much more difficult to follow, if we skip several steps:
\begin{align*}
(A\cap B)\triangle (A\cap C)
&= ((A\cap B)\cup(A\cap C))\setminus((A\cap B)\cap(A\cap C)) \\
&= (A\cap(B\cup C)) \setminus (A\cap B\cap C) \\
&= A\cap ((B\cup C)\setminus (B\cap C)) \\
&= A\cap (B\triangle C).
\end{align*}
The third equality here hides much of what we did before, which includes the steps explicitly using De Morgan's laws.
