1. ## Sets: Direct Proof

Show that $(A \cap B) \cup C = A \cap (B \cup C)$ is false for all sets A, B and C using direct proof.

If we were to prove that the statement is true then we would need to show that if $x \in (A \cap B) \cup C$ then $x \in A \cap (B \cup C)$ and if $x \in A \cap (B \cup C)$ then $x \in (A \cap B) \cup C$

To formalise this we denote the following:

Let P(x) denote the proposition function "$x \in (A \cap B) \cup C$" and Q(x) denote the proposition function "$x \in A \cap (B \cup C)$"

Thus we need to show that $\forall x \left(P(x) \to Q(x)\right)$ and $\forall x \left(Q(x) \to P(x)\right)$

Let us first focus on $\forall x \left(P(x) \to Q(x)\right)$

Because we are using a direct proof let us assume the hypothesis P(x) is true. If we can show that Q(x) is false then the universally quantified statement is false and we have complete our proof.

Since P(x) is assumed to be true then $x \in (A \cap B) \cup C$ can be interpreted as $x \in (A \cap B)$ or $x \in C$.

$x \in (A \cap B) \cup C$ is only true if:

1. $x \in (A \cap B)$ is true and $x \in C$ is true

2. $x \in (A \cap B)$ is true and $x \in C$ is false

3. $x \in (A \cap B)$ is false and $x \in C$ is true

Now Q(x) is interpreted as $x \in A$ and $x \in B \cup C$

Now what...?

2. $\displaystyle (A\cap B)\cup C=A\cap(B\cup C)=A=B=C$ if $\displaystyle A=B=C.$ You're going to need conditions on the three sets in order to make it false. It's certainly not going to hold for all sets!

3. Oh yes, I see, thanks very much Ackbeet!!

4. You're very welcome. Have a good one!

5. Do you mean "false for all sets" or "not true for all sets"?

The first is, as Akbeet said, not a true statement so you cannot prove it.

But the second, that there exist some sets for which $\displaystyle (A\cap B)\cup C\ne A\cap(B\ cup C)$, is true. A counter example is the simplest way to show that- exhibit one of those sets.