# Suppose A,B and C are sets. Prove that if A ⊆ B, then A-C ⊆ B-C

I understand this intuitively.

Let's say there's $x \in A$ and $x \in B$. Then $A - C$ means we're taking away every element in set $C$ from set $A$ and similarly the same for $B - C$.

I understand that even if $C$ and $A$ had the same elements, even then $A-C \subseteq B-C$ holds true (the empty set).

But I can't seem to write a formal proof out for it.

Thanks!

• write $A\backslash C$ for A-C ;) – SAJW Oct 28 '16 at 6:39
• So you want to prove that, if every member of $A$ is a member of $B,$ then every member of $A$ which is not in $C$ is a member of $B$ which is not in $C$? Seems obvious, but writing a formal proof is another story. I've never written (or read) a formal proof of anything in my life, so I can't help you. Anyway, how is anybody going to help you find a formal proof, unless you specify what formal system it has to be in? – bof Oct 28 '16 at 6:48

Let $x \in A-C$, then $x \in A$ and $x \notin C$.

Since $x\in A$, we can conclude that $x \in B$.

Can you see that $x \in B-C$?

$A-C$ is a subset of $B.$ Elements of $A-C$ are not present in $C$. So they belong to set $B-C.$

Let $x \in (A-C)$, by definition of difference

$\implies x \in A \land x \not \in C$

by hypothesis

$\implies x\in B \land x\not \in C$

by definition of difference

$\implies x\in(B-C)$

So, if $A\subseteq B$ then $(A-C) \subseteq (B-C)$ $\blacksquare$