How to determine if this is true or false?

Question

$$\exists x \in X, (P(x) \to Q(x))\hspace{0.2cm} \iff (\exists x \in X, P(x))\to (\exists x \in X, Q(x))$$

$$\forall x \in X, (P(x) \to Q(x))\hspace{0.2cm} \iff (\forall x \in X, P(x))\to (\forall x \in X, Q(x))$$

Really don't know where to start

Answer 1

Use Implication Equivalence and the Laws of Negation and Distribution of Quantifiers.

$$\begin{align} & \exists x \in X: (P(x) \to Q(x)) \\[1ex] \iff & \exists x\in X : (\neg P(x)\vee Q(x)) & \text{Implication Equivalence} \\[1ex] \iff & (\exists x\in X: \neg P(x)) \vee (\exists x\in X:Q(x)) & \text{Distribution: Existential over Disjunction} \\[1ex] \iff & \neg(\forall x \in X: P(x))\vee (\exists x \in X: Q(x)) & \text{Universal Negation} \\[1ex] \iff & (\color{red}{\forall} x \in X:P(x))\to (\exists x\in X: Q(x)) & \text{Implication Equivalence} \end{align}$$

So we must conclude that your given equivalence is not generally true.

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For the other one, beware that the Distribution of Universal over Disjunction itself is not an equivalence.

Answer 2

To try examples with these sentences, you have two predicates $P$ and $Q$. Each member of $X$ can satisfy $P(x)$ or not and (separately) satisfy $Q(x)$ or not. You can let $X$ have four members. Each one will have one of the four possible truth values for $P$ and $Q$. For example, one of them will have $P(x)$ true and $Q(x)$ false. Now test each sentence for a tautology. If one fails in this case you are done. If a sentence passes, you need to think about whether deleting some element of $X$ will render it false.

Answer 3

In general, to show that : $\varphi \leftrightarrow \psi$, we have to provide a proof in some proof system.

To show instead that $\varphi \leftrightarrow \psi$ does not hold, we have to provide a suitable counter-example to one of the conditionals : $\varphi \rightarrow \psi$, $\psi \rightarrow \varphi$.

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For :

$∀x(P(x) → Q(x)) \leftrightarrow (∀xP(x) → ∀xQ(x))$

we can provide a counter-example showing that :

$(∀xP(x) → ∀xQ(x)) \rightarrow ∀x(P(x) → Q(x))$

does not hold.

Consider an interpretation with domain the set $X = \{ 0,1 \}$ and consider as $P(x)$ the formula $(x = 0)$ and as $Q(x)$ the formula $(x > 0)$.

We have that :

$∀x(x = 0) → ∀x(x > 0)$

is true in $X$, because both the antecedent and the consequent are false.

Consider now :

$∀x[(x = 0) → (x > 0)]$;

choosing $0$ as value for $x$, we have that $(0 = 0) → (0 > 0)$ is false (antecedent true, while consequent false) and thus it is false that $(x = 0) → (x > 0)$ is true for any value of $x$.

Thus $∀x[(x = 0) → (x > 0)]$ is false in $X$.

In conclusion, we have found an interpretation such that $∀xP(x) → ∀xQ(x)$ is true and $∀x(P(x) → Q(x))$ is false, i.e. :

$∀xP(x) → ∀xQ(x) \nvDash ∀x(P(x) → Q(x))$,

and thus :

$\nvDash (∀xP(x) → ∀xQ(x)) \rightarrow ∀x(P(x) → Q(x))$.

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The same interpretation can be used to show that :

$∃x(P(x) → Q(x)) \nvDash ∃xP(x) → ∃xQ(x)$.

Consider now as $P(x)$ the formula $(x > 0)$ and as $Q(x)$ the formula $(x < 0)$.

Now we have that $(x > 0) \rightarrow (x < 0)$ is true for $0$ as value of $x$, and thus $∃x[(x > 0) \rightarrow (x < 0)]$ is true in $X$.

But $∃x(x > 0)$ is true while $∃x(x < 0)$ is false, and thus $∃x(x > 0) \rightarrow ∃x(x < 0)$ is false.