logic equivalence question?

Question

I am trying to prove how these 2 formulas are equivalent using logical equivalence laws.

(p ↔ q) → r = (¬p∧q)∨(¬q∧p)∨r

For (p ↔ q) → r, What I tried to do was

(p → q) ∧ (q → p) V r (Bi conditional law)

(p v ¬q) ∧ (q V ¬p) v r (conditional law)

(¬q v p) ∧ (¬p v q) v r (commutative law)

From there I got stuck as I wasn't sure for the next equivalence law.

What would be next?

Thanks.

Answer 1

You made a small mistake as after the Conditional Law (and DeMorgan) you should get:

$$(p \land ¬q) \lor (q \land ¬p) \lor r $$

That is, you get:

$$((p \rightarrow q) \land (q \rightarrow p)) \rightarrow r \Leftrightarrow \text{ (Conditional Law)}$$

$$\neg ((\neg p \lor q) \land (\neg q \lor p)) \lor r \Leftrightarrow \text{ (DeMorgan)}$$

$$\neg (\neg p \lor q) \lor \neg (\neg q \lor p)) \lor r \Leftrightarrow \text{ (DeMorgan x 2)}$$

$$(p \land \neg q) \lor (q \land \neg p) \lor r \Leftrightarrow \text{ (Commutation)}$$

$$(q \land \neg p) \lor(p \land \neg q) \lor r \Leftrightarrow \text{ (Commutation x 2)}$$

$$(\neg p \land q) \lor(\neg q \land p) \lor r$$