Show that An(B-C)=(AnB)-(AnC)

1) Let x be an arbitrary element in A $\cap$ (B-C)

$\Rightarrow$ x $\in$ A and x $\in$ (B-C)

$\Rightarrow$ x $\in$ A, and x $\in$ B and x $\notin$ C

In either case, this implies x $\in$ (A and B) but x $\notin$   (A and C)

$\Rightarrow$ x $\in$ (A$\cap$B) - (A$\cap$C)

Hence, A $\cap$(B-C) $\subset$ (A$\cap$B) - (A$\cap$C) ------- (1)

2) Again let y be an arbitrary element in (A$\cap$B) - (A$\cap$C)

   y   $\in$(A$\cap$B) but y  $\notin$ (A$\cap$C)

$\Rightarrow$ y $\in$A, and y $\in$  B but y $\in$ A and y  $\notin$ C

$\Rightarrow$ y $\in$ A $\cap$(B-C)

So, (A$\cap$B) - (A$\cap$C) $\subset$ A $\cap$(B-C)   ---- (2)

Thus from (1) & (2), it is proved that

A $\cap$(B-C) = (A$\cap$B) - (A$\cap$C)