In relation to classic semantics, let $ \alpha $ be an arbitrary of $ \mathcal L_1 ^ {=} $ - closed formula ...

Question

In relation to classic semantics, let $ \alpha $ be an arbitrary of $ \mathcal L_1 ^ {=} $ - closed formula, such that $ \vDash \alpha $ and $ \Delta $ a set of $ \mathcal L_1 ^ { =} $ - closed formula, such that the formula $ \lnot \alpha $ belongs to $ \Delta $, so $\Delta $$\vDash $$\alpha $ is valid?

Is the answer no, right?

Because let's assume that $ \Delta $ = {$ \beta, \phi, P \to Q, \lnot \alpha, .... $} this means that all formulas are true, so $ \lnot \alpha $ is true, so if $ \alpha $ is true and is in $ \Delta $, then $ \lnot \alpha $ cannot be in $ \Delta $, but by chance $ \lnot \alpha $ is in $ \Delta $, so $ \alpha $ is not a semantic consequence (sorry I don't know how to write the symbol of this).

Answer 1

Your intuition that "$\neg \alpha$ cannot be in $\Delta$" goes in the right direction: If $\alpha$ is valid, then any set of formulas containing $\neg \alpha$ is unsatisfiable (aka contradictory or inconsistent), i.e. there can be no interpretation that makes all formulas in $\Delta$ true.

But precisely because $\Delta$ is unsatisfiable, it entails $\alpha$: Remember that $\Delta \vDash \alpha$ iff every interpretation that satisfies all formulas in $\Delta$ also satisfies $\alpha$, i.o.w., there is no interpretation that satisfies all formulas in $\Delta$ but fails $\alpha$. Now since there is no interpretation that can satisfy $\Delta$ to begin with, then in particular there can be no such counter interpretation which invalidates the entailment: $\Delta \vDash \alpha$ is vacuously valid because $\Delta$ is unsatisfiable.

The inference is actually valid for a second reason: There can be no interpretation that satisfies all formulas in $\Delta$ and simulateneously does not satisfy $\alpha$, because there is no interpretation at all that would not satisfy $\alpha$ -- after all, we assumed that $\alpha$ is valid. In classical logic, adding additional premises to an already valid formula never invalidates the entailment: $\Delta \vDash \alpha$ is trivially valid because $\alpha$ is valid.

In general, we have that

• $\Gamma \vDash \phi$ for arbitary formulas $\phi$, if $\Gamma$ is unsatisfiable, and
• $\Gamma \vDash \phi$ for arbitrary sets of formulas $\Gamma$, if $\phi$ is valid.

Answer 2

I'm assuming you meant to write that $\lnot\alpha$ is in $\Delta$, right?

The statement is true. Since $\models\alpha$, every structure satisfies $\alpha$. Thus every model of $\Delta$ is a model of $\alpha$, which is the definition of $\Delta\models\alpha$.

The fact that $\lnot\alpha\in \Delta$ is irrelevant here - but it also allows us to conclude that $\Delta$ has no models at all!