Absurd argument showing every first-order theory is consistent

Question

Could someone point out to me where the error(s) are in my reasoning below?

First-order logic (with the sequent calculus as the proof system) is both sound and complete. This is the same as saying the set $S$ of semantic consequences of a theory (the statements that are true in every model for which the statements of the theory are also true) is the same as the set $T$ of its syntactic consequences (using the sequent calculus). This is because soundness means that $T$ is a subset of $S$ and completeness means that $S$ is a subset of $T$, i.e. that $S = T$.

This means that every theory is consistent: it is clear that $S$ cannot contain both a statement and its negation (since a model can only satisfy one of the two), but by the above the same must be true for $T$, so that the theory is consistent.

A theory is a collection of specific mathematical axioms? If so, the inconsistency can be there. — Mauro ALLEGRANZA, Oct 25 '23 at 07:56
If a set of axioms is inconsistent, then by explosion the set $S$ of semantic consequences of the theory is equal to the set of all formulas of the language. — Mauro ALLEGRANZA, Oct 25 '23 at 07:58
@MauroALLEGRANZA I agree with your last comment, but I don't see why your first two comments are true (except for the first sentence of the first one). — Abced Decba, Oct 25 '23 at 08:06
I believe your argument shows that every theory which has a model is consistent. — user14111, Oct 25 '23 at 08:14
soundness means: if the axioms are true in some model M, then also the theorems are so. If the set of mathematical axioms is inconsistent, then they have no models: this does not contradict soundness of the underlying logic. — Mauro ALLEGRANZA, Oct 25 '23 at 08:16
See Sound and Complete and Relationship between consistency, strong completeness and soundness. — Mauro ALLEGRANZA, Oct 25 '23 at 08:21
See also Is consistency the same as sound or complete, in mathematical logic? — Mauro ALLEGRANZA, Oct 25 '23 at 08:22

Alex Kruckman · Accepted Answer · 2023-10-25T15:03:43.463

The mistake in your argument occurs (as usual) when you write "it is clear...".

"it is clear that $S$ cannot contain both a statement and its negation (since a model can only satisfy one of the two)"

Here, to conclude that $S$ cannot contain both $\varphi$ and $\lnot \varphi$, you implicitly assume that your theory has a model.

Recall that for a theory $\Sigma$, the set $S$ of semantic consequences of $\Sigma$ is $$S = \{\varphi\mid \forall M\,(M\models \Sigma\to M\models \varphi)\}.$$ If $\Sigma$ has no models, then for every sentence $\varphi$, $\varphi\in S$ (vacuously). In particular, both $\varphi$ and $\lnot \varphi$ are in $S$, and hence also in $T$. So $\Sigma$ is inconsistent.

What the argument above shows is that if $\Sigma$ has no models, then $\Sigma$ is (syntactically) inconsistent. The contrapositive is that every consistent theory has a model - which, incidentally, is the main way the completeness theorem is used in model theory.

The converse, that an inconsistent theory has no models, is clear. So we have that the consistent theories are exactly those with no models (and the consistent theories are exactly those with models).

Ahh I missed that assumption! – Abced Decba Oct 25 '23 at 15:08 — Abced Decba, Oct 25 '23 at 15:08

Absurd argument showing every first-order theory is consistent

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