Answer:
The following proof makes use of logic equivalences.
Step-by-step explanation:
We have to show that [tex]x\in A/(A\cap B)[/tex] is equivalent to say that [tex]x\in (A\cup B)/B[/tex]. In fact, if [tex]x\in A/(A\cap B)[/tex], by definition this is equivalent to say that [tex]x\in A\, \text{and}\, x\notin A\cap B[/tex], this is equivalent to say that[tex]x\in A\, \text{and}\, (x\notin A\, \text{or}\, x\notin B) [/tex], this is equivalent to [tex] x\in A\, \text{and}\, x\notin B[/tex], and this is equivalent to say that [tex](x\in A\, \text{or} \, x\in B)\, \text{and}\, x\notin B[/tex], and this is equivalent to say that [tex]x\in (A\cup B)/B[/tex].
The Following image can be useful.