Prove that for all sets $A$ and $B$, $( A \cup B ) - B = A$ implies $A \cap B =\varnothing$. In the following proof we harness properties such as:
For all set $A$,
(i) $A = A - \varnothing$
(ii) $A \cap\varnothing=\varnothing$
Proof:
Let $A$ and $B$ be arbitrary sets and suppose $( A \cup B ) - B = A$.
Given (i) then $$( A \cup B ) - B = A - \varnothing.$$
So then, $( A \cup B ) = A$ and $B = \varnothing$.
That being so,
$$A \cap B = ( A \cup B ) \cap \varnothing = \varnothing$$
If we apply (ii) to the foregoing equivalence relation. Therefore
$$A \cap B = \varnothing$$
Is this proof right? if it is I would like to see alternative ones.
Suppose that $x \in A \cap B$ existed. Then $x \in A$ but also $x \in A \cup B$ and $x \in B$ so that $x \notin (A \cup B) - B$. This contradicts the starting assumption that $A=(A\cup B)-B$, so $A \cap B = \emptyset$.
Suppose $(A\cup B) \setminus B = A$.
Let $x \in A\cap B$.
Then $x \in A$ and $x\in B$.  Then $x \in B$ so $x\not \in M\setminus B$ for any set $M$ so $x\not \in (A\cup B) \setminus B$.  So $x \not \in A$. But that's a contradiction.
So if $(A\cup B)\setminus B =A$ then $x \in A\cap B$ is impossible.  So $A\cap B=\emptyset$.
Not this in no way means $B$ is empty; just $A$ and $B$ are dijoint.
.....
Alternatively.  If $(A\cup B)\setminus B = A$ then if $x \in A$ then $x\in (A\cup B) \setminus B$ so $x \not \in B$.  So $x$ in both $A$ and $B$ is impossible so $A\cap B=\emptyset.
Or....
If $(A\cup B)\setminus B = A$ and if $x \in B$ then $x\not \in (A\cup B)\setminus B = A$ so $x$ in both $B$ and $A$ is impossible.
....
An alternative way would be to prove that $(M\cup N)\setminus N = M\setminus N$ always.  I'll leave that proof to you.
That means $(A\cup B) \setminus B= A\setminus B= A$.  So if $x\in B$ the $x$ is removed for $A$ to get $A\setminus B$ so $x\not \in A$.
Etc.