Animated Solution for Mathematics - Sets and Relations: The statement $A\to (B\to A)$ is equivalent to:

Welcome, future engineer! Today, we are going to peel back the layers of a classic logical puzzle. You might look at the statement $A\to (B\to A)$ and feel a bit intimidated by the nested arrows.

But I want you to see this not as a wall of symbols, but as a beautiful, symmetric structure waiting to be simplified. Logic is the bedrock of computer science and mathematics, and mastering these manipulations is your first step toward thinking like a true architect of systems.

The first thing we must do is demystify the implication arrow. In logic, the statement $P\to Q$ is not just a mysterious arrow; it is mathematically equivalent to $\sim P\vee Q$.

This is our most powerful tool. It allows us to trade the rigid, directional nature of an implication for the flexible, commutative nature of an 'OR' statement.

Let's apply this to our inner bracket, $(B\to A)$. By treating $B$ as our $P$ and $A$ as our $Q$, the expression transforms into $\sim B\vee A$. Now, our original statement $A\to (B\to A)$ becomes:

Now, we have an outer implication to deal with. We apply the same identity again. Here, our $P$ is $A$, and our $Q$ is the entire bracket $(\sim B\vee A)$.

Negating the first term gives us $\sim A$, and the arrow becomes an 'OR' operator. So, we have $\sim A\vee (\sim B\vee A)$.

This is where the magic happens. Because the 'OR' operator is associative, the placement of the parentheses does not matter. We can rewrite this as:

Look closely at the term $(\sim A\vee A)$. This is the Law of Excluded Middle. A statement is either true or false; it cannot be both, and it cannot be neither.

Therefore, the disjunction of a statement and its negation is always true. We call this a tautology, denoted by $t$. Our expression is now $\sim B\vee t$.

In logic, any statement ORed with a tautology is itself a tautology. Because one part of an 'OR' statement is always true, the entire statement is true, regardless of the value of $B$. Thus, our original expression is a tautology.

To find the equivalent expression, we simply need to find which option also simplifies to a tautology. Let's look at Option 2: $A\to (A\vee B)$.

Applying our identity, we get $\sim A\vee (A\vee B)$. Using the associative law again, we get $(\sim A\vee A)\vee B$, which simplifies to $t\vee B$, and finally, $t$.

Since both the original statement and Option 2 simplify to a tautology, they are logically equivalent. You have just navigated the heart of propositional logic. Keep this elegance in mind—whenever you see complex implications, look for the path to a tautology!