We are given two propositional formulas:

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

We need to prove that these two formulas are logically equivalent. To do this, we will simplify both formulas using logical rules and show that they reduce to the same form.

Formula 1: $(\neg P \iff Q) \implies R$

We'll begin by breaking down this formula step by step:

Step 1: Expand $\neg P \iff Q$

The biconditional $\iff$ can be rewritten as two implications:

$\neg P \iff Q \equiv (\neg P \implies Q) \land (Q \implies \neg P)$

Step 2: Rewrite each implication

Using the implication identity $A \implies B \equiv \neg A \lor B$, we rewrite both implications:

$\neg P \implies Q \equiv P \lor Q$ $Q \implies \neg P \equiv \neg Q \lor \neg P$

So, the biconditional becomes:

$\neg P \iff Q \equiv (P \lor Q) \land (\neg Q \lor \neg P)$

Step 3: Substitution into the original formula

Now we substitute this back into the original formula:

$((P \lor Q) \land (\neg Q \lor \neg P)) \implies R$

Step 4: Apply implication

Again, using the implication identity $A \implies B \equiv \neg A \lor B$, we rewrite the entire formula as:

$\neg ((P \lor Q) \land (\neg Q \lor \neg P)) \lor R$

Step 5: Apply De Morgan’s Law

To simplify $\neg ((P \lor Q) \land (\neg Q \lor \neg P))$, we use De Morgan's Law:

$\neg ((P \lor Q) \land (\neg Q \lor \neg P)) \equiv \neg (P \lor Q) \lor \neg (\neg Q \lor \neg P)$

Using De Morgan's Law again on both parts:

$\neg (P \lor Q) \equiv \neg P \land \neg Q$ $\neg (\neg Q \lor \neg P) \equiv Q \land P$

Thus, the formula simplifies to:

$(\neg P \land \neg Q) \lor (P \land Q) \lor R$

Formula 2: $(P \land Q) \lor (\neg P \land \neg Q) \lor R$