Animated Solution for Mathematics - Sets and Relations: Let $\Delta ,\nabla \in \{\wedge ,\vee \}$ be such that $(p\to q)\Delta (p\nabla q)$ is a tautology. Then

We begin with the expression $(p \rightarrow q) \Delta (p abla q)$. The first hurdle is the implication $p\to q$.

Many students memorize this as a table, but it is best viewed as a region in a Venn diagram. The implication $p\to q$ is logically equivalent to $\sim p\vee q$.

Imagine two circles, $P$ and $Q$. The region $\sim p$ is everything outside circle $P$. When we take the union with $Q$, we are essentially shading everything except the 'crescent' of $P$ that does not overlap with $Q$. This is our starting territory.

Our goal is to make the entire expression a tautology, which in the world of sets, is the Universal Set $U$. We need to choose our operators $\Delta$ and $ abla$ to ensure that no corner of our logical space is left unshaded.

Let us test the hypothesis where $ abla = \vee$. By choosing the union for the second part, $(p\vee q)$, we are covering both circles $P$ and $Q$ entirely.

Now, we must decide how to combine our first region $(\sim p\vee q)$ with this new region $(p\vee q)$ using the operator $\Delta$.

If we choose $\Delta =\vee$, we are looking at the union of these two regions:

Because the operator is the same throughout, we can invoke the Associative and Commutative laws. We are free to rearrange the terms as we please:

Look at that! The term $(\sim p\vee p)$ is the classic Law of Excluded Middle. It is the logical certainty that $p$ must be either true or false, meaning $(\sim p\vee p)\equiv T$.

Simultaneously, the Idempotent Law tells us that $(q\vee q)\equiv q$. Our expression has collapsed into $T\vee q$.

We are left with $T\vee q$. In the realm of logic, the Domination Law states that if you have a True statement joined by an 'OR' ($\vee$) to any other statement, the result is always True.

It does not matter if $q$ is true or false; the $T$ has already won the day. The entire expression is a tautology.

We have successfully navigated the logic, verified it through algebra, and arrived at the conclusion that $\Delta =\vee$ and $ abla = \vee$.

Remember, in the exam hall, do not just calculate—visualize. When you see these logical operators, see the Venn diagrams, see the regions of truth, and let the laws of algebra guide you to the answer.