if $\vDash \alpha \ \Rightarrow \ \vDash \beta$ then $\vDash \alpha \to \beta$?

Question

if $\vDash \alpha \ \Rightarrow \ \vDash \beta$ then $\vDash \alpha \to \beta$ ?

Is this proposition true? And what about converse? I also wonder about the difference between $\Rightarrow$ and $\to$ in this proposition.

Answer 1

$\to$ is the propositional connective "if..., then...". It is part of the language of propositional logic and it is used (with other connectives) to symbolize the formulas of the formal system.

The symbol $\vDash \alpha$ is not part of the language of the system but is part of the meta-language : it asserts the fact that formula $\alpha$ is a tautology (or logically valid).

Thus, $\Rightarrow$ is used in te meta-language to abbreviate "if..., then..." when expressing properties of the formal system.

---

Regarding the question :

if $⊨ α \ ⇒ \ ⊨ β$, then $⊨ α → β$ ?

the answer is no.

Consider the atom $p$ as $\alpha$ and the atom $q$ as $\beta$.

We have that $\vDash p$ is $\text {False}$.

Thus :

$\vDash p \ \Rightarrow \ \vDash q$ holds,

because $\text {False } \Rightarrow \text { False}$ is $\text {True}$.

But $\nvDash (p \to q)$, because $p \to q$ is not a tautology.

---

About the converse : assume that $\vDash \alpha \to \beta$ and $\vDash \alpha$.

We reason by contradiction, i.e. assume that $\nvDash \beta$. This means that for some truth assignment $v$ we have $v(\beta)=$ f.

But from $\vDash \alpha$, we have that $v(\alpha)=$ t.

Thus :

$v(\alpha \to \beta)=$ f,

contradicting $\vDash \alpha \to \beta$.