The logical consequence of an empty set of premises.

Question

I am studying propositional logic by self-study, using a dutch book. I hope I am translating the terms to the correct English term. If my words are confusing, please please just let me know instead of editing my question (feel free to edit if I don't respond in a few weeks or so).

The logical consequence from an empty set of premises (a tautology) confuses me. My book explains the logical consequence $\Sigma \vDash \psi$ as such:

If every interpretation that is a model for $\Sigma$, is also a model for $\psi$, then $\psi$ is the logical consequence of $\Sigma$.

If $\Sigma$ is an empty set, my reasoning is as follows: there exists no interpretation that makes $\Sigma$ true, there is no model for $\Sigma$, so there exist no models that are true for $\Sigma$ but not for $\psi$.

Unfortunately, my reasoning seems incorrect as my book explains (translation mine),

If $\Sigma = \emptyset$, then every interpretation is a model for $\Sigma$,

which of course results in $\psi$ being a tautology.

Wikipedia stays closer to what I understand: ($\psi$ is a logical consequence of $\Sigma$) if and only if there is no model in which all members of $\Sigma$ are true and $\psi$ is false. Or, in other words, the set of the interpretations that make all members of $\Sigma$ true is a subset of the set of the interpretations that make $\psi$ true.

Why is every interpretation a model for an empty set of premises? A model is an interpretation that makes a premise true. Nothing is made true (or false) when the set is empty. What am I missing?

Some related material not answering my question (so that this question is not closed too quickly...): Logical consequence (math SE) and Visualing (...) entailment.

Answer 1

See Herbert Enderton, A Mathematical Introduction to Logic (2nd - 2001), page 23 :

take the special case in which $\Sigma$ is the empty set $\emptyset$.

Observe that it is vacuously true that any truth assignment satisfies every member of $\emptyset$. (How could this fail? Only if there was some unsatisfied member of $\emptyset$, which is absurd.) Hence we are left with : $\emptyset \vDash \psi$ iff every truth assignment (for the sentence symbols in $\psi$) satisfies $\psi$.

In this case we say that $\psi$ is a tautology (written $\vDash \psi$).

As highlighted by Enderton, we have a case of vacuous truth.

The condition for an interpretation $I$ to be a model of $\Sigma$ is that all the sentences in $\Sigma$ must be satisfied by the interpretation $I$.

This is :

for all $\sigma$, if $\sigma \in \Sigma$, then $I$ satisfy $\sigma$.

Thus, vacuous truth applies : there are no $\sigma \in \emptyset$.

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What we are trying to convince ourselves is that : if $\emptyset \vDash \psi$, then $\psi$ is a tautology.

Consider again the definition of logically implies; there is a double conditional in place.

Saying that $\Sigma \vDash \psi$ means :

for every interpretation $I$, [ if for every sentence $\sigma$, ( if $\sigma \in \Sigma$, then $I$ satisfy $\sigma$ ), then $I$ satisfy $\psi$ ].

In semi-formal way :

$\forall I [ \forall \sigma(\sigma \in \Sigma \rightarrow I \vDash \sigma) \rightarrow I \vDash \psi ]$.

When we put $\emptyset$ in place of $\Sigma$, the antecedent of the "inner" conditional is false; thus, by truth-table for $\rightarrow$, the conditional is true (and this says nothing about the truth-value of the consequent !).

In this way, the antecedent of the "outer" conditional is true. But we are asserting the fact that $\emptyset \vDash \psi$, i.e. that the "outer" conditional is true.

If it is true and if its antecedent is true, there is only one possibility left : the consequent is true.

I.e.

$I \vDash \psi$.

This hold for every $I$, and thus we can conclude that $\psi$ is a tautology.

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New addition

We can try with another approach.

According to the definition of logical consequence, $\varphi \vdash \psi$ iff $\varphi \land \lnot \psi$ is always false.

Now, assuming that $\Gamma$ is a finite set of sentences, i.e. $\Gamma = \{ \gamma_1, \ldots, \gamma_n \}$, we have that :

$\Gamma \vdash \psi$ iff $\gamma_1 \land \ldots \land \gamma_n \land \lnot \psi$ is always false.

Thus, if $\Gamma = \emptyset$, the above condition boils down to :

$\emptyset \vDash \psi$ iff $\lnot \psi$ is always false.

Obviously, $\lnot \psi$ is always false iff $\psi$ is a tautology.