How to prove if $C \subseteq B$ then $A - B \subseteq A-C$

Question

if $C \subseteq B$ then $A - B \subseteq A-C$

I know $C \subseteq B$ translates to

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

And $A - B \subseteq A-C$ is

$\forall x (x \in A \land x \notin B \implies x \in A \land x \notin C)$

But I'm not sure what the next step is to prove it.

Answer 1

Suppose $x \in A-B$, then it must be that $x\in A$ and $x \notin B$. But, since $C \subseteq B$, $x \notin B$ implies that $x \notin C$.

Answer 2

x is in C implies x is in B. Thus x is not in B implies x is not in C (for all x). In particular this is true for x in A. That is, x is in A and not in B implies x is in A and not in C.

Answer 3

Suppose that $C \subseteq B$.

To show that $A - B \subseteq A - C$, we have to show that whenever $x \in A - B$, then $x \in A - C$.

Let $x \in A - B$. This means $x \in A$ and $x\notin B$.

Now, $C \subseteq B$, so $x \notin B \implies x \notin C$. (The best way to convince yourself of this, is to draw a little diagram representing $B$ and $C$, makking sure $C$ is completely inside $B$, and then seeing it graphically. More formally, this statement is the contrapositive of $C \subseteq B$).

Hence, $x\in A$ and $x \notin C$, hence $x \in A - C$.

Therefore, $A - B \subseteq A-C$. On your request, I can symbolize the above argument completely.

Answer 4

We are given that all Courageous people are Bold. (Hypothesis: $C\subset B$))

When from a particular set of people A, we had thrown away the bold persons, ($A-B$) we had thrown away the courageous too. They will be from A but not courageous, that is they will be part of $A-C$.