We are tasked with proving that the following two propositional formulas are logically equivalent:

1. $(\neg P \iff Q) \implies R$
2. $(P \land Q) \lor (\neg P \land \neg Q) \lor R$

Steps for the proof:

We will use logical equivalences and transformations to demonstrate that both formulas yield the same truth values for any assignment of $P$, $Q$, and $R$.

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Step 1: Simplifying the first formula

Let's begin with the first formula: $(\neg P \iff Q) \implies R$

Recall that $A \iff B$ is equivalent to $(A \implies B) \land (B \implies A)$. So, we can rewrite $\neg P \iff Q$ as: $(\neg P \implies Q) \land (Q \implies \neg P)$ Thus, the first formula becomes: $[(\neg P \implies Q) \land (Q \implies \neg P)] \implies R$

Now, simplify the implications:

• $\neg P \implies Q$ is equivalent to $P \lor Q$ (since $\neg A \implies B \equiv A \lor B$).
• $Q \implies \neg P$ is equivalent to $\neg Q \lor \neg P$.

So, we rewrite the expression as: $[(P \lor Q) \land (\neg Q \lor \neg P)] \implies R$

Step 2: Analyzing the first formula

Now, observe that $(P \lor Q) \land (\neg Q \lor \neg P)$ is actually the definition of the exclusive-or (XOR) equivalence of $P$ and $Q$: $(P \lor Q) \land (\neg Q \lor \neg P) \equiv P \iff Q$ Thus, the first formula can now be rewritten as: $(P \iff Q) \implies R$

This means "if $P$ and $Q$ are equivalent, then $R$ must be true."

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Step 3: Simplifying the second formula

Now, consider the second formula: $(P \land Q) \lor (\neg P \land \neg Q) \lor R$

The first two terms, $(P \land Q) \lor (\neg P \land \neg Q)$, represent the equivalence of $P$ and $Q$ (since they are true when $P$ and $Q$ are either both true or both false). This is exactly the definition of $P \iff Q$.

Thus, the second formula simplifies to: $(P \iff Q) \lor R$

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Step 4: Comparing the two formulas

We now have the following simplified versions of the two formulas:

1. $(P \iff Q) \implies R$
2. $(P \iff Q) \lor R$

Now, let's check the equivalence of these two formulas.

• $(P \iff Q) \implies R$ is equivalent to $\neg (P \iff Q) \lor R$ (using the implication equivalence $A \implies B \equiv \neg A \lor B$).
• This simplifies to $(\neg (P \iff Q)) \lor R$.

The second formula, $(P \iff Q) \lor R$, is exactly the same, since $\neg (P \iff Q)$ means $P$ and $Q$ are not equivalent, but either way, $R$ must hold.

Thus, both formulas are logically equivalent: $(P \iff Q) \implies R \equiv (P \iff Q) \lor R$

Conclusion:

The two formulas $(\neg P \iff Q) \implies R$ and $(P \land Q) \lor (\neg P \land \neg Q) \lor R$ are logically equivalent.

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Would you like more details or have any further questions on specific parts of this proof?

Here are 5 related questions to explore further:

1. How can you derive the equivalence $\neg P \implies Q \equiv P \lor Q$?
2. What is the truth table for the logical biconditional $P \iff Q$?
3. How do De Morgan’s laws apply in simplifying logical formulas?
4. Can logical equivalences be used to prove the consistency of more complex logical systems?
5. How does the structure of implication $\implies$ differ from biconditional $\iff$ in terms of truth values?

Tip: Always start by simplifying implications and biconditionals using known logical equivalences. This will make the structure of the formulas clearer!