Semantic Consequence Union Negation

Theorem
Let $$U$$ be a set of logical formulas.

Let $$P$$ be a logical formula.

Let $$U \models P$$ denote that $$U$$ is a logical consequence $$P$$.

Then:
 * $$U \models P$$

iff:
 * $$U \cup \left\{{\neg P}\right\}$$ has no models.

Sufficient Condition
Suppose that $$U \models P$$.

Let $$\mathcal M$$ be a model such that:
 * $$\mathcal M \models U \cup \left\{{\neg P}\right\}$$

Then from Logical Consequence with Union, we have that:
 * $$\mathcal M \models U$$

and by definition of logical consequence:
 * $$\mathcal M \models P$$

So we have that $$\mathcal M \models P$$ and $$\mathcal M \models \neg P$$

By definition of contradiction, it follows that there can be no such model.

So $$U \models P$$ is a sufficient condition for $$U \cup \left\{{\neg P}\right\}$$ to have no models.

Necessary Condition
Suppose that $$U \cup \left\{{\neg P}\right\}$$ has no models.

There are the following possibilities:


 * $$\neg P$$ has no models, in which case it is itself a contradiction.

Hence from Tautology and Contradiction, $$P$$ is a tautology.

Hence $$P$$ is true under every model.

Hence whatever $$U$$ may be, every model of $$U$$ is also a model of $$P$$:
 * $$U \models P$$


 * $$U$$ has no models, in which case every model of $$U$$ is also a model of $$P$$ vacuously, and so:
 * $$U \models P$$


 * There exists at least one model of $$U$$, but each one does not model $$\neg P$$.

Let $$\mathcal M$$ be such a model.

We have that:
 * $$\mathcal M \not \models \neg P$$

so by definition of negation:
 * $$\mathcal M \models P$$

Thus $$\mathcal M \models U$$ implies that $$\mathcal M \models P$$.

Hence by definitinon:
 * $$U \models P$$

In all cases $$U \models P$$.

So $$U \models P$$ is a necessary condition for $$U \cup \left\{{\neg P}\right\}$$ to have no models.