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.
 A: 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 :


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


