In which direction is this statement true?

Question

Having a hard time how I would go at this question.

$\exists x \in G, P(x) \wedge Q(x) \iff (\exists x \in G, P(x)) \wedge (\exists x \in G, Q(x))$

$\forall x \in D, P(x) \vee Q(x) \iff (\forall x \in D, P(x)) \vee (\forall x \in D, Q(x))$

Answer 1

Hint 1: Note that in the first bi-conditional, RHS is true whenever LHS is true, since if there is a single thing with both properties P and Q (LHS)---say a Pink Queen, then it does follow that there is something Pink and that there is a Queen. If this is a material bi-conditional, then even when LHS is false or impossible the '$\Longrightarrow$' direction will be true. Consequently the '$\Longrightarrow$' direction of the first formula is always true (valid). The '$\Longleftarrow$' direction isn't valid however, for suppose there is something Pink and there is a thing that is a Queen, but the Pink thing and the thing that is a Queen needn't be the same thing---the thing that is a Queen may be White (say), and as such not Pink. Try to use similar reasoning for the second bi-conditional.

Hint 2: You may want to construct a counter-model for either of the directions. Here two-item models will suffice, as suggested by $\color{navy}{\text{columbus8myhw}}$ above. Let $G=D=\{ a,b\}$, and recall that for finite domains $D$, $\exists x \in D, P(x)$ is equivalent to $P(a_1) \lor P(a_2) \lor...\lor P(a_n)$ for an n sized domain, and $\forall x \in D, P(x)$ is equivalent to $P(a_1) \land P(a_2) \land...\land P(a_n)$. This gives you a way to transform those FOL frormulae into PC formulae, the validity of which you can check using semantic trees or truth tables, if no obvious countermodels come to mind.

Hint 3: The first bi-conditional for a two element model would be equivalent to the formula $( P(a) \land Q(a)) \lor ( P(b) \land Q(b)) \iff ( P(a) \lor P(b)) \land ( Q(a) \lor Q(b))$. Now consider the model where $P=\{a \}$ and $Q=\{b \}$, and note that this is a counter-model for the '$\Longleftarrow$' direction, since the RHS conjunction is true since both of its conjuncts are true, i.e. $P(a)$ and $Q(b)$ is true on this model, and LHS disjunction isn't true since both of its disjuncts fail to be true on account of there being an absence of an object $\alpha \in G$ such that both $P(\alpha)$ and $Q(\alpha)$.

The above counter-model is merely a formalization of the earlier counterexample given in Hint 1: think of $a$ as the Pink thing in $G$, and $b$ as the Queen in $G$, but that doesn't entail that there needs to be a single object in $G$ that is both a Queen and Pink. Hope that helps.