Question: Let $a,b,c,d>0$ and $x,y,z,t < 0$ real numbers. Prove that at least one of the inequalities $ax+bc > 0$, $xy+cz > 0$, $yd+zt > 0$ or $da+tb>0$ is false.

I assume by contradiction they are all true, but what to do next?

Answer: 
As already pointed out in the comments, $A$ and $B$ are sets, not statements.

Still, there are obvious connections between sets and logical statements. For example, the set $A \cap B$ is the set of all objects that are elements of $A$ *and* of $B$. Or, to make the connection between $\cap$ and $\land$ even more clear:

$A \cap B = \{x \mid x \in A \land x \in B \}$

Using this format, we can also say:

$(A \cap B') \cup (B \cap A') = \{ x \mid (x \in A \land \neg (x \in B)) \lor (\neg (x \in A) \land x \in B) \}$

If we can use the logical biconditional, you can simplify this to:

$(A \cap B') \cup (B \cap A') = \{ x | (x \in A \leftrightarrow \neg (x \in B) \}$
For any set $S$, there is a predicate $P\_S$ such that $x\in S\iff P\_S(x)$ (in other words, $S=\{x:P\_S(x)\}$). For instance, if we are working in the domain of integers we can assign the set of even numbers to the predicate $E(n):=2\mid n$. If $S\_E$ is the set of even numbers, then $n\in S\_E$ has exactly the same meaning as $E(n)$. In some theories, this is actually the *definition* of a predicate/set.$^1$

Using this correspondence, it is possible to convert set-theoretic statements into logical ones and vice-versa. In this sense sets *can* be interpreted as [equivalence classes of] statements, provided that certain rules are observed.

There are many different ways to do this depending on the particular set-theory and logic you choose, the choice of logical connectives, and the presence or absence of quantifiers.

For propositional logic,$^2$ I would use the following:

Logical negation $\neg$ corresponds to the complement of a set $'$.

Logical disjunction $\lor$ corresponds to the union of sets $\cup$.

Logical conjunction $\land$ corresponds to the intersection of sets $\cap$.

Predicate application $P\_X(x)$ corresponds to set membership $x \in X$.

For the example provided, Let $P\_A$ and $P\_B$ be predicates such that $a\in A\iff P\_A(a)$ and $b\in B\iff P\_B(b)$. Then...

$$x\in(A\cap B')\cup (B\cap A')\iff((P\_A\land\neg P\_B)\lor(P\_B\land\neg P\_A))(x)$$

You can simplify the logical statement using *exclusive disjunction* to $P\_A\oplus P\_B$. The corresponding set is the *symmetric difference* $A\triangle B=(A\cup B)\setminus(A\cap B)$.

In general, if two statements are logically equivalent (for example $P\land Q\equiv\neg (\neg P\lor\neg Q)$, then the corresponding sets will be equivalent as well ($S\_P\cap S\_Q=(S\_P'\cup S\_Q')'$). The connection between sets and logical statement in general is an important topic in the study of *Boolean algebras*.

Hugely helpful Hasse diagram: <https://commons.wikimedia.org/wiki/File:Logical_connectives_Hasse_diagram.svg>

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Notes for nit-pickers:

$^1$ Actually, it is more common to refer to predicates [up to logical equivalence] as *classes* rather than sets - particularly in axiomatic set theory where the identification of predicates with sets is the same as *unrestricted comprehension*, which leads to paradoxes like *Russel's Paradox*.

$^2$ Technically, this is predicate logic. However, it is much closer to propositional logic than true first-order logic in its application. In an alternative formulation, you could assign a proposition $\varphi\_S$ to each set $S$ using a formula $\varphi\_S:=\forall x.P\_S(x)$ for the appropriate predicate $P\_S$. In this case, I would say that the set $S$ is the domain/universe in which $\varphi\_S$ is *true*.