Prove statement with a counter-example: For all sets S, T and V, V ∪ (S ∩ T) = (V ∪ S) ∩ T is false Title says it all.
Prove the statement is false with a counter-example: 
For all sets S, T and V, V ∪ (S ∩ T) = (V ∪ S) ∩ T
Explain thoroughly if possible. I'm new to this
 A: Hint: Think simple. Try letting one of $V$, $S$ and $T$ be the empty set $\varnothing$ and come up with a counter-example.
A: You need to find sets S, T and V such that that equality is false. 
The motivation behind the question is the fact that $(V \cup S) \cap T = (V\cap T) \cup (S \cap T)$, and that the positioning of the brackets does matter. Notice from that, that for the equality to be false we require $V\neq V \cap T$ (so $V$ must contain elements that $T$ does not). It should be straightforward to find an example with all sets only having 2 or 3 elements.
A: One way finding a counterexample is to try and simplify $\;V \cup (S \cap T) = (V \cup S) \cap T\;$.  Taking the simplest possible approach, viz. expanding the definitions and applying logic, you could calculate as follows:
\begin{align}
& V \cup (S \cap T) \;=\; (V \cup S) \cap T \\
\equiv & \;\;\;\;\;\text{"definition of set equality, i.e., extensionality"} \\
& \langle \forall x :: x \in V \cup (S \cap T) \;\equiv\; x \in (V \cup S) \cap T \rangle \\
\equiv & \;\;\;\;\;\text{"definitions of $\;\cup\;$ and $\;\cap\;$"} \\
& \langle \forall x :: x \in V \lor (x \in S \land x \in T) \;\equiv\; (x \in V \lor x \in S) \land x \in T \rangle \\
\equiv & \;\;\;\;\;\text{"logic: in left hand side, distribute $\;\lor\;$ over $\;\land\;$ -- the only thing we can do"} \\
& \langle \forall x :: (x \in V \lor x \in S) \land (x \in V \lor x \in T) \;\equiv\; (x \in V \lor x \in S) \land x \in T \rangle \\
\equiv & \;\;\;\;\;\text{"logic: extract common conjunct from both sides of $\;\equiv\;$"} \\
& \langle \forall x :: x \in V \lor x \in S \;\Rightarrow\; (x \in V \lor x \in T \;\equiv\; x \in T) \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify consequent of $\;\Rightarrow\;$"} \\
& \langle \forall x :: x \in V \lor x \in S \;\Rightarrow\; (x \in V \;\Rightarrow\; x \in T) \rangle \\
\equiv & \;\;\;\;\;\text{"logic: merge antecedents"} \\
& \langle \forall x :: (x \in V \lor x \in S) \land x \in V \;\Rightarrow\; x \in T \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify antecedent"} \\
& \langle \forall x :: x \in V \;\Rightarrow\; x \in T \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\subseteq\;$"} \\
& V \subseteq T \\
\end{align}
Now you should be able to find a counterexample.
A: This is a simple counter - example:
suppose you have:
$$S=\left\{a\right\},V=\left\{b\right\},T=\left\{c\right\}$$
$$V\cup(S\cap T)=\left\{b\right\}$$
while:
$$(V\cup S)\cap T=\emptyset$$
