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

Welcome, fellow traveler of the mathematical mind. Today, we are not just solving a problem; we are embarking on a journey into the very bedrock of logical reasoning. We are looking at the concept of a tautology—a statement that stands as an eternal truth, regardless of the variables we plug into it.

It is the logical equivalent of a universal constant. We are given two mysterious operators, $\Delta$ and $ abla$, chosen from the set $\{\wedge ,\vee \}$. We are told that the implication $p abla q \Rightarrow ((p \Delta q) abla r)$ is a tautology.

In logic, $A\Rightarrow B$ is a promise. It says, 'If $A$ is true, then $B$ must be true.'

The only way to break this promise—to make the statement False—is to have a scenario where $A$ is True, but $B$ is False. To prove our expression is a tautology, we must act like a detective and try to find a counterexample. If we cannot find one, we have our proof.

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Let's start by testing $ abla = \wedge$. Our expression becomes:

Can we break this? Let's try setting $p=T$ and $q=T$. The premise $(p\wedge q)$ becomes True.

Now, look at the conclusion: $((T\Delta T)\wedge r)$. Regardless of what $\Delta$ is, $(T\Delta T)$ will evaluate to True. So the conclusion simplifies to $T\wedge r$, which is just $r$. If we choose $r=F$, the conclusion becomes False.

We have found our counterexample: $p=T,q=T,r=F$ makes the premise True and the conclusion False. Thus, $T\Rightarrow F$ is False. This proves that $ abla$ cannot be $\wedge$. It must be $\vee$!

Now that we know $ abla = \vee$, our expression updates to:

We need to find $\Delta$. Let's test $\Delta =\wedge$ first. The expression becomes:

Let's try to break this. If we set $p=T$ and $q=F$, the premise $(p\vee q)$ is True. But look at the conclusion: $((T\wedge F)\vee r)\equiv (F\vee r)\equiv r$.

If we set $r=F$, the conclusion is False. Again, we have $T\Rightarrow F$, which is False. So $\Delta$ cannot be $\wedge$.

This leaves us with only one option: $\Delta =\vee$. Let's verify this. The expression becomes:

If we let $X=p\vee q$, we have $X\Rightarrow (X\vee r)$. If $X$ is True, then $X\vee r$ is definitely True, and $T\Rightarrow T$ is True. If $X$ is False, then $F\Rightarrow (F\vee r)$ is $F\Rightarrow r$, which is always True.

It works! We have found our operators: $ abla = \vee$ and $\Delta =\vee$.

Now, we turn to the target expression: $(p abla q) \Delta r$. Substituting our findings, we get:

By the Associative and Commutative Laws of logic, we can rearrange this as:

Since $\Delta =\vee$, we can rewrite this as $(p\Delta r)\vee q$. This matches the required logical form perfectly. You have the tools to solve any logical puzzle—keep practicing, and keep questioning!