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

Solution 1:

Hint: Think simple. Try letting one of $V$, $S$ and $T$ be the empty set $\varnothing$ and come up with a counter-example.

Solution 2:

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.

Solution 3:

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.