Predicate logic proof Prove the following formula.
$$ \vdash (\exists x)(A \land B) \lor (\exists x)(A \land C) \equiv (\exists x)(A \land (B \lor C))$$
The question is number 10 in chapter 6 in "Mathematical Logic" by Tourlakis.
My try :
$$ (\exists x)(A \land B) \lor (\exists x)(A \land C) $$
$$ A \land (\exists x)B \lor A \land (\exists x)C $$
$$ A \land ((\exists x)B \lor (\exists x)C)$$
I got stuck in that step.

You must follow Git Gud's answer and complete the proof with the formal details according to theorems and rules of your textbook.
We must use 6.1.7 Theorem. (Distributivity of $\forall$ over $\land$) : $\vdash (\forall x)(A \land B) \equiv (\forall x) A \land (\forall x) B$, page 158.
Start with : $(\exists x)(A \land (B \lor C)$ and rewrite with $\forall$ :

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

then use De Morgan : $\lnot (\forall x) (\lnot A \lor \lnot (B \lor C))$, De Morgan again : $\lnot (\forall x) (\lnot A \lor (\lnot B \land \lnot C))$, distribute : $\lnot (\forall x) ((\lnot A \lor \lnot B) \land (\lnot A \lor \lnot C))$, and De Morgan again : $\lnot [(\forall x) (\lnot (A \land B) \land \lnot (A \land C))]$.
Now we apply Th 6.1.7 :

$\lnot [(\forall x) \lnot (A \land B) \land (\forall x)\lnot (A \land C)]$.

Now, we "switch" again from $\forall$ to $\exists$ :

$\lnot [\lnot (\exists x) (A \land B) \land \lnot (\exists x) (A \land C)]$.

Finally, we apply again De Morgan :

$\lnot \lnot [(\exists x) (A \land B) \lor (\exists x) (A \land C)]$

and Double Negation :


$(\exists x) (A \land B) \lor (\exists x) (A \land C)$.


