In this problem, A(x) and B(x) are two predicates over the same domain.

Question

For each of the statements below, state whether it is true or false. If true, explain why each of its directions → and ← is true. If false state which direction is false and give a counterexample.

(a) ∀x (A(x) ∨ B(x)) ↔ ∀xA(x) ∨ ∀xB(x)

(b) ∀x (A(x) ∧ B(x)) ↔ ∀xA(x) ∧ ∀xB(x)

(c) ∃x (A(x) ∨ B(x)) ↔ ∃xA(x) ∨ ∃xB(x)

(d) ∃x (A(x) ∧ B(x)) ↔ ∃xA(x) ∧ ∃xB(x)

I am stuck on this question. From reasoning it out via sentences, I believe that the correct (or true) arrows for each of the statements are:

(a) Left

(b) Left

(c) Right

(d) Both

How should I be approaching this problem?

Answer 1

For b and c both directions hold true ...

b: if everything has both properties $A$ and $B$, then obviously everything has property $A$, and everything has property $B$, ... and vice versa.

c: if there is something that has either property $A$ or $B$, then if that thing has property $A$, then there is something with property $A$, and if that thing has property $B$, then there is something with property $B$. So, either there is something with property $A$, or there is something with property $B$

Going the other way around: Assume that either there is something with property $A$, or something with property $B$. Well, in both cases there would then be something woth either property $A$ or $B$

For a, you are right: only the left direction holds: assume that either everything has property $A$, or everything has property $B$. In both cases, everything has either property $A$ or property $B$. For a counterexample against going right: take as the domain all integers, and let $A(x)$ be '$x$ is even', and let $B(x)$ be '$x$ is odd'

For d, going to the right holds: if there is something with both properties $A$ and $B$, then gclearly there is something (namely that very thing) with property $A$, and there is something (again, that very thing) that has property $B$. For a counterexample going left, see above counterexample.

Answer 2

How should I be approaching this problem?

You should have a reason for the answers you have given. You just need to express your rational.

Eg: you have stated you believe $∀x\,(A(x) ∨ B(x)) ↔ ∀x\,A(x) ∨ ∀x\,B(x)$ is only valid for a Left conditional ($\gets$).   However, why do you think this is so?

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If all are one or all are the other, then anything is one or the other.

However, it is possible to have all be one or the other without everything being one, or everything being the other.

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The first can be demonstrated with a simple proof by cases, but you were not asked for this.

The second can be justified with a counter example. An interpretation where the conditional is not true. vis:

Consider the domain of natural numbers with $A$ interpreted as "is odd" and $B$ as "is even". Well every natural number is either odd or even, but not all natural numbers are odd and not all natural numbers are even.
So $\forall x~(A(x)\vee B(x)$ does not entail that $\forall x~A(x)\vee\forall x~B(x)$.