Show, by the element method that, for all subsets P, Q, and R of U, (P − Q) ∩ (R − Q) = (P ∩ R) − Q. i do not seem to what element method is. Please explain me how to use the element method
 A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\notag \\ #1 \quad & \quad \text{"#2"} \notag \\ \quad & }
\newcommand{\endcalc}{\notag \end{align}}
$The element method, also known as element chasing, is about proving two sets $\;A\;$ and $\;B\;$ equal by proving they have exactly the same element, i.e., proving that -- for any $\;x\;$ -- if $\;x \in A\;$ then also $\;x \in B\;$, and vice versa.  Very often this just comes down to expanding the definitions and simplifying.
In this specific case, the simplest thing is to start at the most complex side, so $\;(P - Q) \cap (R - Q)\;$, and calculate which $\;x\;$ this set contains:
$$\calc
x \in (P - Q) \cap (R - Q)
\calcop{\equiv}{definition of $\;\cap\;$}
x \in P - Q \;\land\; x \in R - Q
\calcop{\equiv}{definition of $\;-\;$, twice}
(x \in P \land x \not\in Q) \;\land\; (x \in R \land x \not\in Q)
\calcop{\equiv}{logic: simplify by factoring out the common conjunct $\;x \not\in Q\;$}
\ldots
\endcalc$$
A: Element method is
Assume x as an element of p-q intersection r-q now prove that the same x is an element of p intersection r minus q.
For each x this should be applicable.
