Prove difference between a set and the union of other two sets

Question

This question asked which of the two following formulas was always right whilst the other one was sometimes wrong, and what was the necessary and sufficient condition for the formula which was sometimes incorrect to be always right:

• (i) $\; A - (B - C) = (A - B) \cup C $

• (ii) $\; A - (B \cup C) = (A - B) - C $

Using Venn diagrams, I guessed (ii) was always right, and (i) would be wrong if $A$ and $C$ are disjoint. Thus, the necessary and sufficient condition for (i) to be correct was $A$ and $C$ contain some elements in common.

My questions are:

1. Were my guesses correct?

2. How could I prove the two formulas above using words/expressions rather than using Venn diagrams?

Thanks a lot.

Answer 1

Yes, your guesses are correct (except for “the necessary and sufficient condition for (i) to be correct” — I explain it in my other answer.)

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Ad (i):

To prove, that it is not always true, it is sufficient to provide particular sets $A$, $B$, $C$ which don't satisfy the equation (i.e. so called contra example).

So let $A$, $B$, $C$ are as in this table, where are progressively calculated expressions from them:

Row	Expression	Value
1	$$A$$	$$\{1\}$$
2	$$B$$	$$\emptyset$$
3	$$C$$	$$\{2\}$$
4	$$(B-C)$$	$$\emptyset$$
5	$$A−(B−C)$$	$$\{1\}$$
6	$$(A−B)$$	$$\{1\}$$
7	$$(A−B)∪C$$	$$\{1, 2\}$$

By comparing rows 5 and 7 we see that the equation is not satisfied.

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Ad (ii):

To prove the equality for every sets $A$, $B$, $C$, we need to prove that every element in the set on the left-hand side of the equality

$$A−(B∪C)=(A−B)−C$$

belongs to the right-hand set, and vice versa.

So let

$$x \in A−(B∪C).$$

Then by definition of the set subtraction

$$x \in A, \ \text{but}\ x \notin (B \cup C)$$

But if $x \notin (B \cup C)$, then by definition of the set union

$$\text{neither}\ x \in B, \ \text{nor}\ x \in C$$

which mean — again by definiton of the set subtraction — that

$$x \in (A - B), \ \text{ but }\ x \notin C$$

i.e. — once more by definiton of the set subtraction — that

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

So we proved that the arbitrary element $x$ of the left-hand side belongs to the right-hand side, too.

The reversed statement we may prove by similar manner.

Answer 2

To obtain

the necessary and sufficient condition for the formula

(i) $\; A - (B - C) = (A - B) \cup C $

which was sometimes incorrect to be always right

let's first use Venn diagrams:
 

As we may see, the results are different. For an easy comparison let's display and enumerate individual parts of every result:

From the picture follows that for obtaining the same results, the parts 4 and 5 must be empty.

But what represent parts 4 and 5 together? The set $C - A$, of course.

It means that this set must be empty, i.e. (from definition of set subtraction) there must not be an element in $C$ which is in the same time not in $A$.

In other words, every element of the set $C$ must belong to the set $A$, too, which means that the set $C$ must be a subset of the set $A$.

So we may state our hypothesis:

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The necessary and sufficient condition for the formula

$$A - (B - C) = (A - B) \cup \tag 1C$$

to be always correct is

$$C \subset A.\tag 2$$

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Now we prove it:

1. Sufficiency.

   Let $C \subset A.\ $ Then:

   1. The left-hand side of $(1)$ is a subset of its right-hand side, because:

      If $x \in A−(B−C)$,

      then $x \in A$, but $x \notin (B−C)$. From the definition of the set subtraction follows that either $x \notin B$, or $x \in C$.

      1. If $x \notin B$, then $x \in (A-B)$, so indeed $x \in (A - B) \cup C$.

      2. If $x \in C$, then $x \in (A - B) \cup C$, too.

      So if $x$ belongs to the left-side hand of $(1)$, it belongs to the right-side hand of $(1)$, too, which means that $$A - (B - C) \subset (A - B) \cup C \tag 3$$

   2. The left-hand side of $(1)$ is a superset of its right-hand side, because:

      If $x \in (A - B) \cup C$,

      then $x \in (A - B)$, or $x \in C$.

      1. If $x \in (A - B)$, then $x \in A$, but $x \notin B$. But if $x \notin B$, $x \notin (B-C)$, too. So $x \in A - (B - C)$.

      2. If $x \in C$, then (by presumption $(2)$) $x \in A$, too, but certainly it not belongs to $B-C$. So, again, $x \in A - (B - C)$.

      So if $x$ belongs to the right-side hand of $(1)$, it belongs to the left-hand side hand of $(1)$, too, which means that $$A - (B - C) \supset (A - B) \cup C \tag 4$$

   From $(3)$ and $(4)$ follows, that

   $$A - (B - C) = (A - B) \cup C,$$

   so the $C \subset A$ is the sufficient condition for this equation.

    

2. Necessity.

   Let $A - (B - C) = (A - B) \cup C,$

   and let $x \in C$. Then $x$ belongs to the right-hand side of this equation, which means that it belongs to the left-hand side of this equation, too, which in turn means that $x$ certainly belongs to $A$, too.

   So $C \subset A$, so we prove it as a necessary condition for $(1)$.