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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.

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Out Of Bounds
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1 Answers1

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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.

$$¬(∀x)(¬A ∧ ¬B)$$

$$¬((∀x)¬A ∧ (∀x)¬B)$$

$$¬(¬(\exists x)A ∧ ¬(\exists x)B)$$

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

Hunan Rostomyan
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