If no interpretations satisfy a set of formulae U, is it possible for $U\models A$?

Question

Note: '$ \models$' denotes logical consequence, defined as

If $U \models A$, then $A$ is a logical consequence of $U$, if and only if every interpretation that satisfies U also satisfies $A$, for a set of formulae $U$ and a formula $A$.

My question is, if no interpretation satisfies $U$, can $A$ be a logical consequence of $U$?

Example:

Prove or Disprove: ${(A \land B), \lnot B} \models \lnot A$

In that example, there does not exist any interpretations that satisfy both $A\land B$ and $\lnot B$. So does that mean it is disproven?

Reference: Ben-Ari Mathematical Logic for Computer Science.

Answer 1

1. "There does not exist any interpretations that satisfy both $A \lor B$ and $\neg B$". Wrong. Suppose $A$ is true and $B$ is false.

2. To answer your general question, however. $\Gamma \vDash \varphi$ holds, by definition, iff there is no interpretation which makes every sentence in $\Gamma$ true and $\varphi$ false. In the special case where there is no interpretation which makes everything in $\Gamma$ true, then it will also be the case, trivially, that there is no interpretation which makes everything in $\Gamma$ true and $\varphi$ false, whatever $\varphi$ is. And hence it will be true that $\Gamma \vDash \varphi$, whatever $\varphi$ is.