Prove or disprove (A - B)' = A' U B

Question

How do I prove such a statement? Do I begin from the LHS and manipulate the expression until I get to the RHS? I know that to show that two sets are equal I need to show that the left is a subset of B and vice versa -- do I have to do this for the LHS and RHS?

As for actually simplifying the expressions, I do not know how to start...

Also, I was told using Venn diagrams to show both give the same shaded area is not a valid method of proof.

Any help will be greatly appreciated, thanks in advance.

Answer 1

By D Morgan's Law and the definition of 'difference of set $B$ from $A$' we get $(A - B)^{'} $ = $(A \cap B')' $ = $(A' \cup B) $.

Answer 2

Let $x $ be any element of the universal set. There are exactly 4 possibilities:

1) $x $ is in neither $A$ nor $B $.

Then $x $ is not in $A-B\subset A $. So $x$ is in $(A-B)'$. And $x $ is not in $A $ so $x $ is in $A'\subset A'\cup B $.

So $x $ is in both $(A-B)'$ and $A'\cup B $.

[Note: this all follows from $x $ not being in $A $. It does not matter if $x$ is or is not in $B $.]

2) $x $ is not in $A $ but in $B $.

The result above holds (if $x $ is not in $A $ it is in both set whether or not it is in $B $)

3) $x$ in $A $ but not in $B $.

Then $x $ is in $A-B $. So $x $ is not in $(A-B)'$.

$x $ is not in $A'$ and $x $ is not in $B $ so $x $ is not in $A'\cup B $.

So $x $ is in neither set.

4) $x $ is in both $A $ and $B $.

Then $x $ is in $B $ so $x $ is not in $A-B $. So $x $ is in $(A-B)'$. And $x $ is in $B $ so $x $ is in $A'\cup B $.

So $(A-B)'$ and $A'\cup B $ both have the same elements and exclude the same elements. Which means the are equal.

Answer 3

Like you say, a way of proving that the sets $X$ and $Y$ are equal is to show $X \subseteq Y$ and $Y\subseteq X$. To do this show first that for an arbitrary $x \in X$ we also have that $x \in Y$ and for arbitrary $y \in Y$, we have that $y \in X$.

Now to prove your specific question.

Show $(A\backslash B)' = A'\cap B$.

First let $A,B$ be subsets of a set $X$. Let $x \in (A \backslash B)'$, then we have that $ x \not \in A \backslash B$. So $x \in X \backslash (A \backslash B)$ which means $x \in (X\backslash A) \cup B = A'\cup B$. Thus $(A\backslash B)' \subseteq A' \cup B$.

Now let $x \in A' \cup B$, so consider two cases. If $x \in B$, then $x \not \in A \backslash B$. So $x \in (A \backslash B)'$. If $x \in A'$ then $x \not \in A$ so $x \not \in A \backslash B$. So $x \in (A \backslash B)'$. In either case $x \in (A \backslash B)'$ so $A' \cup B \subseteq (A \backslash B)'$.

We draw the conclusion that $(A\backslash B)' = A'\cap B$.