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This is what I have done thus far:

Theorem. $(A \cap B ) = (A \cup B) \Longleftrightarrow A = B$

Proof.

$A \cup B \Longrightarrow A = B$

Assume $A \cup B$ is true.

Given $A = B$ we know that:

$$A \cup B \Longleftrightarrow (A \subseteq B) \wedge (B \subseteq A)$$

$$\Longleftrightarrow (x \in A \Leftrightarrow x \in B)$$

$A = B \Longrightarrow A \cup B$

Assume $A = B$ is true.

Given $A \cup B$ let $x \in (A \cup B)$

So, $x \in A \vee x \in B$

I am unsure of how to proceed, and I am thrown off by the fact that there are three consecutive statements. Am I on the right track? If not, then how can I prove this?

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$$A\cap B\subseteq A,B\subseteq A\cup B=A\cap B,$$ hence all these people are equal.

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    I am having trouble coming up with a way to come up with the proof for this. I have shown $A\cap B\subseteq A$ in another proof, but I do not know how to apply it in this case. – 2017-02-24
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    Look at the inclusions, you finally have $A\cap B\subseteq A\subseteq A\cap B$, so $A= A\cap B$, and similarly for $B$. Isn't that clear? – 2017-02-24
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Proof: $A∪B⟹A=B$

...

$A=B⟹A∪B$

No.   $A\cup B$ is a set, not a predicate.

I am unsure of how to proceed, and I am thrown off by the fact that there are three consecutive statements.

There are not three consecutive statement. There is an equivalence between two equalities.

You want to prove $~\Big( (A\cap B)=(A\cup B)\Big) ~\iff~ A=B~$ via proving

  • $\Big( (A\cap B)=(A\cup B)\Big) ~\implies~ A=B$
  • $A=B~\implies~\Big( (A\cap B)=(A\cup B) \Big)$

You do the former by showing that if the intersection equals the union then every element that is in $A$ must be in $B$, and by symmetry that all in $B$ must be in $A$. (IE: $A\subseteq B$ and $B\subseteq A$.)

Then you demonstrate the converse.