# Transform $(\phi \vee \psi) \wedge (\neg \phi \vee \neg \psi)$ to $(\phi \wedge \neg \psi) \vee (\neg \phi \wedge \psi)$ using equivalences?

$$(\phi \oplus \psi) \equiv (\phi \vee \psi) \wedge (\neg \phi \vee \neg \psi) \equiv (\phi \wedge \neg \psi) \vee (\neg \phi \wedge \psi)$$

I found the first one in a book and thought of the second one myself and was under the impression that I can transform one into the other using just the usual equivalences for classical propositional logic. How do I transform $$(\phi \vee \psi) \wedge (\neg \phi \vee \neg \psi)$$ to $$(\phi \wedge \neg \psi) \vee (\neg \phi \wedge \psi)$$ using just the usual equivalences?

$$(\phi \vee \psi) \wedge (\neg \phi \vee \neg \psi)$$
$$\equiv (\phi \wedge (\neg \phi \vee \neg \psi)) \vee (\psi \wedge (\neg \phi \vee \neg \psi))$$
$$\equiv (\phi \wedge \neg \psi) \vee (\neg \phi \wedge \psi)$$
XOR is a negation of equivalence: $$\neg (P\iff Q)\equiv \neg\underbrace{(P\implies Q)\land(Q\implies P)}_{\text{conjunction}}\equiv\\\neg(P\implies Q)\lor \neg(Q\implies P)\equiv (P\land\neg Q)\lor(Q\land\neg P)$$