Notation for Tautologies

Question

I've been stuck for a while in this question and so far I don't understand the flaw of my reasoning please if you guys could help me out.

See, this is my context. From the definition of argument we have that it is valid if and only if there's no way that being the premises all true the conclusion be false. In symbols:

Model theory: if $\Gamma$ is a set of premises and $\alpha$ the conclusion then $\Gamma\models\alpha$.

Proof theory: if $\Gamma$ is a set of premises and $\alpha$ the conclusion then $\Gamma\vdash \alpha$.

If $\Gamma$ is for example $\{\alpha_1,\alpha_2,...,\alpha_n\}$ then we write $\alpha_1,\alpha_2,...,\alpha_n\models\alpha$ (or respectively $\vdash$).

Now my problem. In most books I see they define $\models\alpha$ to represent the fact that $\alpha$ is a tautology. If I think of $\vdash\alpha$ as being a theorem deduced simply by the axioms of deduction then it makes sense because I don't need actually any premises and hence they're always true because they are deduced from the axioms of deduction which turn out to be tautologies. But what about in terms of proof theory, meaning is it true that $\models\alpha$ if and only if $\emptyset\models\alpha$? intuitively I think is should, the thing is that I don't know how to prove it. Let's see my attempt, which I'm going to write by using logic symbols for the sake of simplicity:

Definition: $\Gamma\models \alpha$ if and only if for all valuation $v$, $\forall x(x\in\Gamma([x]_v=1))\longrightarrow[\alpha]_v=1$.

If $\Gamma=\emptyset$ then the expression: $(x\in\Gamma([x]_v=1))\longrightarrow[\alpha]_v=1$ is always true because $\forall x(x\notin\emptyset)$ and then $x\in\Gamma([x]_v=1)$ is false. But this does not mean that $\alpha$ is a tautology.

Answer 1

You seem to be confused about the notion of a vacuous truth.

If I say "all mermaids are charming", it is vacuously true because there are no mermaids in real life (if you say it is false, then there must be a mermaid that isn't charming).

Concerning your question, recall that, from the definition of semantic consequence, what '$\emptyset \vDash \alpha$' says is that for any interpretation $v$ and $\phi \in \emptyset$, if $v(\phi)=1$ then $v(\alpha)=1$. As you know, it is a particular case of $\Gamma \vDash \alpha$ when $\Gamma=\emptyset$. When is convenient, sometimes we abbreviate it to $\vDash \alpha$. Hence, their relation is not something we can prove, simply because $\vDash \alpha$ has no meaning by itself. (Of course informally we can always write it, but one has always $\emptyset \vDash \alpha$ in mind istead.)

Now $\emptyset$ has no elements $\phi$ at all, so that the antecedent holds for all interpretation $v$. Naturally, if $\alpha$ is true in any interpretation then it is a valid formula (tautology).

Answer 2

Lets define $\models\alpha$ as a shorthand of $\emptyset\models\alpha$ and we will proof that $\models\alpha$ if and only if $\alpha$ is a tautology. This implies taking $\models \alpha$ to mean "$\alpha$ is tautology" is an equivalent definition.

Given an arbitrary valuation $v$ there are four cases:

1) $\forall x\in\emptyset([x]_v=0)$

2) $\forall x\in\emptyset([x]_v=1)$

3) $\exists x\in \emptyset([x]_v=0)$

4) $\exists x\in \emptyset([x]_v=1)$

Also if $\varphi(x)$ is any formula then

$\forall x\in\emptyset(\varphi(x)):=\forall x(x\in \emptyset\longrightarrow\varphi(x))$

and

$\exists x\in \emptyset(\varphi(x)):=\exists x(x\in \emptyset\wedge \varphi(x))$

then $1)$ and $2)$ are always true and $3)$ and $4)$ are always false because $x\in\emptyset$ is false.

Now if $v$ is any valuation and $\emptyset\models \alpha$, case 1) as such adds nothing directly to what we already know: $[\alpha]_v=1$ or $[\alpha]_v=0$, both options are possible and this doesn't implies $\alpha$ is a tautology. BUT, if $v$ is any valuation and $\emptyset\models\alpha$, case $2)$ implies $[\alpha]_v=1$ by definition of "$\models$" and therefore $\alpha$ must be a tautology.

Conversely, if $\alpha$ is a tautology then if $v$ is any valuation such that $\forall x\in \emptyset([x]_v=1)$ then trivially $[\alpha]_v=1$ and therefore $\models\alpha$.