Existential Quantifier Distributivity in First Order Logic

Question

It is well-known that the following formula holds for boolean functions involving free variables: $$\exists x,P(x)\wedge Q(x)\neq (\exists x_1,P(x_1))\wedge(\exists x_2,Q(x_2)),$$ $$\exists x,P(x) \vee Q(x)=(\exists x_1,P(x_1)) \vee (\exists x_2,Q(x_2)).$$ However, I wonder whether the following holds: $$\exists x,P(x)\wedge Q(x)= (\exists x_1,P(x_1))\wedge(\exists x_2,Q(x_2))\wedge (x_1=x_2)?$$

Answer 1

You wonder whether the following formula holds: $$\exists x(P(x)\wedge Q(x)) \quad\leftrightarrow\quad (\exists x_1 P(x_1) \wedge \exists x_2 Q(x_2) \wedge x_1=x_2).\tag1$$ Note that $(1)$ is equivalent to (all the small letters are variables) $$\exists x(P(x)\wedge Q(x)) \quad\leftrightarrow\quad (\exists y P(y) \wedge \exists z Q(z) \wedge p=q).\tag1$$ This assignment and symbolisation key shows that the open formula $(1)$ does not have a definite truth value:

$$P(x) \;:=\; x>1\\ Q(x) \;:=\; x<3\\ p \;:=\; 5\\ q_1 \;:=\; 5\\ q_2 \;:=\; 6$$

On the other hand, the open formula $$\exists x(P(x)\wedge Q(y)) \quad\leftrightarrow\quad \exists x P(x)\wedge Q(y) \tag2$$ is valid.

Answer 2

It looks a bit weird, because you introduce $x_1$ and $x_2$ separately from the statement $(x_1=x_2)$. Something better would be to write: $$\exists x, P(x) \wedge Q(x)= \exists x_1, \exists x_2, P(x_1) \wedge Q(x_2) \wedge (x_1=x_2)$$