# Set Intersection is not Cancellable

## Theorem

That is, for a given $A, B, C \subseteq S$ for some $S$, it is not always the case that:

$A \cap B = A \cap C \implies B = C$

## Proof

Let $S = \left\{ {a, b, c}\right\}$.

Let:

$A = \left\{ {a}\right\}$
$B = \left\{ {a, b}\right\}$
$C = \left\{ {a, c}\right\}$

Then:

 $\displaystyle A \cap B$ $=$ $\displaystyle \left\{ {a}\right\}$ $\displaystyle$ $=$ $\displaystyle A \cap C$

but:

 $\displaystyle B$ $=$ $\displaystyle \left\{ {a, b}\right\}$ $\displaystyle$ $\ne$ $\displaystyle \left\{ {a, c}\right\}$ $\displaystyle$ $=$ $\displaystyle C$

$\blacksquare$