0
$\begingroup$

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

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?

$\endgroup$
0
$\begingroup$

I wrote down my question and instantly knew the answer. I just leave this here since I already typed it.

$(\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)) $

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.