3
$\begingroup$

Give an equational proof $$ \vdash (\exists x)(A \lor B) \equiv (\exists x)A \lor (\exists x)B $$


What I tried

$(\exists x)(A \lor B)$

Applying Definition of $\exists$

$\lnot (\forall x)\lnot (A \lor B)$

Applying De morgan

$\lnot (\forall x) (\lnot A \land \lnot B)$

Applying Distributivity of $\forall$ over $\land$

$\lnot (\forall x) \lnot A \land \lnot(\forall x) \lnot B$

Applying Definition of $\exists$

$(\exists x)A \land (\exists x)B$

What can I do next ?

See George Tourlakis, Mathematical Logic (2008) or this post for a list of axioms and theorems.

$\endgroup$
2
$\begingroup$

The ∀-distributivity over ∧ step is wrong. Negation doesn't apply to (∀x) because (∀x) is not a sentence, so you can't just push ¬(∀x) in. Here is one way of proceeding.

$$¬(∀x)(¬A ∧ ¬B)$$

$$¬((∀x)¬A ∧ (∀x)¬B)$$

$$¬(¬(\exists x)A ∧ ¬(\exists x)B)$$

$$(\exists x)A \lor (\exists x)B$$

$\endgroup$

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.