Does $\exists$ distribute over an implication? ie. Is $\exists x \in \mathbb{R}, (p(x) \rightarrow q(x))$ logically equivalent to $(\exists x \in \mathbb{R}, p(x)) \rightarrow (\exists x \in \mathbb{R}, q(x))$. If so, can you give an example of $p(x)$ and $q(x)$ to demonstrate?
Thanks very much for any help in advance.
On
Check it in a finite domain: {x, y}. Let p(x) = a, q(x) = b, p(y) = c, q(y) = d. (Notice a,b,c,d are true/false values). Then just check:
$$((a \implies b) \lor (c \implies d)) \implies (a \lor c) \implies (b \lor d)$$
A counter example is b and d false, c false, a true.
On
The sentence $\exists x[p(x)\to q(x)]$ is equivalent to $\forall xp(x)\to\exists xq(x)$. To see this, use the distributivity of $\exists$ over disjunction: $$\exists x[p(x)\to q(x)]\iff\exists x[\neg p(x)\lor q(x)]\iff\exists x\neg p(x)\lor\exists xq(x)\iff\neg\forall xp(x)\lor\exists xq(x)\iff\forall xp(x)\to\exists xq(x)$$
It' easy to see that $\forall xp(x)\to\exists xq(x)$ is not equivalent to $\exists xp(x)\to\exists xq(x);$ just consider an interpretation where $\exists xq(x)$ is false and $\exists xp(x)$ does not have the same truth value as $\forall xp(x),$ i.e., the former is true and the latter false.
Note that $$ \exists x\in\Bbb R,(x=2\to x=x+1)$$ is true (as witnessed by letting $x=0$), whereas $$ (\exists x\in\Bbb R,x=2)\to (\exists x\in\Bbb R, x=x+1)$$ is false.