Animated Solution for Mathematics - Sets and Relations: The statement $(p\Rightarrow q)\vee (p\Rightarrow r)$ is NOT equivalent to:

Mastering Logical Equivalence

Welcome, future engineers. Today, we are not just solving a problem; we are peeling back the layers of logical structure. In the JEE Advanced examination, Mathematical Logic is often treated as a 'scoring' topic, but it is also a place where students lose marks due to carelessness.

We are going to change that. We are going to look at the expression $(p\Rightarrow q)\vee (p\Rightarrow r)$ and learn how to dismantle it with surgical precision.

The Master Key

When you see an implication arrow like $\Rightarrow$, your first instinct should be to translate it. Why? Because implications are 'conditional'—they depend on the state of the antecedent. But logical operators like $\vee$ (OR) and $\wedge$ (AND) are absolute and easier to manipulate.

We use the fundamental Implication Rule:

Think of this as a translation dictionary. It tells us that saying 'if $p$ then $q$' is logically identical to saying 'either $p$ is false, or $q$ is true.' This is the bedrock upon which we will build our solution.

The Simplification

Let us take our given statement: $(p\Rightarrow q)\vee (p\Rightarrow r)$.

Applying our master key, we replace the first implication with $(\sim p\vee q)$ and the second with $(\sim p\vee r)$. Now, our expression looks like this:

Because the entire expression is connected by 'OR' operators, the associative and commutative laws allow us to rearrange this freely. We can group the terms:

Here, we invoke the Idempotent Law: $A\vee A\equiv A$. This is a beautiful, simplifying moment where the two $\sim p$ terms collapse into one. We are left with:

This is our 'Reference Form.' It is the DNA of the original statement. Any option that is logically equivalent to the original must simplify to this exact structure.

The Investigation

Now, we play the role of a detective. We must test each option to see if it shares this DNA.

Consider Option C: $p\Rightarrow (q\vee r)$. Applying the rule, we get $\sim p\vee (q\vee r)$, which is simply $\sim p\vee q\vee r$. It matches! This is equivalent.

Consider Option A: $(p\wedge (\sim r))\Rightarrow q$. Apply the rule: negate the antecedent and keep the consequent. We get $\sim (p\wedge \sim r)\vee q$.

Now, apply De Morgan's Law to the first part. The 'AND' becomes 'OR', and the negation distributes: $(\sim p\vee r)\vee q$. Rearranging, we get $\sim p\vee q\vee r$. It matches again!

Consider Option D: $(p\wedge (\sim q))\Rightarrow r$. Following the same logic, we get $\sim (p\wedge \sim q)\vee r$. Applying De Morgan's Law, we get $(\sim p\vee q)\vee r$, which is $\sim p\vee q\vee r$. Another match!

The Trap

Finally, we arrive at Option B: $(\sim q)\Rightarrow ((\sim r)\vee p)$.

Let us apply the rule. We negate the antecedent $(\sim q)$, which gives us $\sim (\sim q)\equiv q$. Then we keep the consequent $((\sim r)\vee p)$. Putting it together, we get:

Rearranging this, we get $p\vee q\vee \sim r$.

Compare this to our reference form: $\sim p\vee q\vee r$.

Do you see it? The negation is on the wrong variable! In our reference form, $p$ is negated. In Option B, $r$ is negated. They are fundamentally different logical statements. We have found our mismatch.

Logic is not about memorizing truth tables; it is about understanding the flow of truth. By converting everything into a standard 'OR' form, we stripped away the complexity and revealed the underlying structure.

Remember, in the exam, don't rush to draw tables. Use the algebra, trust the laws, and keep your eyes sharp for the placement of those negation signs. You have the tools—now go and conquer.