The following statement (p → q) → [(~p → q) → q] is: (1) a tautology (2) equivalent to ~p → q (3) eq

Q: Wait, why is $p \rightarrow q$ equivalent to $\sim p \vee q$? That feels counter-intuitive.

It is a common point of confusion! Think of $p \rightarrow q$ as a promise: 'If $p$ happens, then $q$ must happen.' The only way to break this promise is if $p$ happens but $q$ does not. In all other cases (if $p$ doesn't happen, or if $q$ happens), the promise is kept. $\sim p \vee q$ captures exactly this: either $p$ is false (the condition didn't trigger) or $q$ is true (the result happened). They are logically identical.

Q: Can I just use a truth table instead of these laws? It feels safer.

You absolutely can, and for a problem with only two variables ($p$ and $q$), a truth table is a perfectly valid and robust method. However, mastering the algebraic laws (De Morgan's, Distributive, etc.) is a superpower. It builds your intuition for symbolic manipulation, which is essential for more complex problems where truth tables become unwieldy.

Q: Why does $\sim q \vee q$ become $T$ (Tautology)?

This is the Law of Excluded Middle. A statement $q$ must be either true or false. Therefore, the statement '$q$ is false OR $q$ is true' covers all possibilities. Since it is always true, we represent it as $T$. When you 'AND' any statement with $T$, you are essentially saying 'Statement AND True', which is just the statement itself.

Question

The following statement $(p \rightarrow q) \rightarrow [(\sim p \rightarrow q) \rightarrow q]$ is:

Options:

(A) a tautology

(B) equivalent to $\sim p \rightarrow q$

(C) equivalent to $p \rightarrow \sim q$

(D) a fallacy

Sigma Percentile · Accepted Answer

### The Architecture of Logic: Demystifying the Tautology

Welcome, fellow traveler in the realm of mathematics. Today, we are going to dismantle a logical expression that, at first glance, looks like a tangled knot of arrows and brackets.

The expression is $(p \rightarrow q) \rightarrow [(\sim p \rightarrow q) \rightarrow q]$. It might seem intimidating, but I want you to take a deep breath.

In logic, as in life, complexity is often just simplicity in disguise. Our goal is to peel back the layers until we find the elegant truth hidden at the core.

### Phase 1: The Bridge of Implication

To navigate this maze, we need a map. Our most powerful tool is the fundamental identity of implication:

$$p \rightarrow q \equiv \sim p \vee q$$

Think of this as a bridge. Whenever you see an implication arrow, you can replace it with a 'not-or' structure.

This is the key that unlocks the entire problem. We are going to apply this rule repeatedly, transforming the expression step by step.

### Phase 2: Peeling the Onion

Let us start from the very center, the innermost bracket: $(\sim p \rightarrow q)$. Using our bridge, this becomes $\sim(\sim p) \vee q$.

Now, recall the law of double negation: the negation of a negation is simply the original statement. So, $\sim(\sim p)$ becomes $p$. Our inner term simplifies beautifully to $p \vee q$.

Now, let us look at the larger square bracket: $(p \vee q) \rightarrow q$. Applying our implication rule again, we get $\sim(p \vee q) \vee q$.

We need to apply De Morgan's Law to that negation. The negation of an 'or' is the 'and' of the negations. Thus, $\sim(p \vee q)$ becomes $(\sim p \wedge \sim q)$.

Our expression is now $(\sim p \wedge \sim q) \vee q$.

### Phase 3: The Distributive Dance

We are now at the heart of the problem. We have $(\sim p \wedge \sim q) \vee q$.

To simplify this, we use the Distributive Law, which works just like multiplying out terms in algebra:

$$(A \wedge B) \vee C \equiv (A \vee C) \wedge (B \vee C)$$

Applying this to our expression, we get $(\sim p \vee q) \wedge (\sim q \vee q)$.

Look closely at the second part: $(\sim q \vee q)$. This is a classic tautology.

A statement $q$ is either true or false, so the disjunction of $q$ and its negation is always true. We denote this as $T$.

So, our expression becomes $(\sim p \vee q) \wedge T$. Since any statement 'and' True is just the statement itself, we are left with $\sim p \vee q$.

### Phase 4: The Final Revelation

Wait a moment! Does $\sim p \vee q$ look familiar? It is exactly our implication rule in reverse!

It is equivalent to $p \rightarrow q$. This means that the entire, intimidating square bracket $[(\sim p \rightarrow q) \rightarrow q]$ is logically equivalent to $p \rightarrow q$.

Now, let us substitute this back into our original expression. We started with $(p \rightarrow q) \rightarrow [(\sim p \rightarrow q) \rightarrow q]$.

With our simplification, this becomes $(p \rightarrow q) \rightarrow (p \rightarrow q)$. This is the form $A \rightarrow A$.

If the premise and the conclusion are identical, the implication is always true. We have arrived at a **Tautology**.

### Conclusion

Look at what we have achieved. We took a complex, intimidating string of symbols and, through the systematic application of logical laws, reduced it to a simple, undeniable truth.

This is the essence of mathematical problem-solving. It is not about memorizing formulas; it is about seeing the structure, trusting the process, and enjoying the journey of simplification.

Keep practicing, keep questioning, and remember: even the most complex problems are just a series of small, manageable steps waiting to be taken.

Animated Solution for Mathematics - Sets and Relations: The following statement $(p \to q) \to [(\sim p \to q) \to q]$ is:

Select Answer:

Visualized Solution

The Logical Statement

The Implication Rule

Simplifying the Innermost Bracket

Expanding the Square Bracket

Applying De Morgan's Law

Using the Distributive Law

The Identity of Negation

Recognizing the Pattern

Conclusion - A Tautology

The Sigma Insight: Types of Sets and Set Operations

The Architecture of Logic

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Animated Solution for Mathematics - Sets and Relations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The following statement (p→q)→[(∼p→q)→q] is: