Why does a finite set having a model imply that the set is consistent

Question

Assuming the soundness theorem to be true, can someone explain why if we assume $\Sigma$ has a model $M$. Then $\Sigma$ is consistent ?

Answer 1

The soundness theorem says that every sentence that you can prove is true about every model. Therefore you can show that something cannot be proved by finding a model where it is false.

Consistency means that you can't prove a contradiction.

A contradiction is never true in any structure, no matter whether it's a model or not.

Thus, if you have a model, then you know that a contradiction will be false in the model, and therefore it cannot be proved from the theory it's a model of. Since the theory cannot prove a contradiction, it is consistent.

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This holds no matter whether $\Sigma$ is finite or not, by the way.

Answer 2

Assume $\Sigma$ has a model $M$, which we write as $M \models \Sigma$. If $\Sigma$ were inconsistent, then $\Sigma \vdash \bot$, but by the soundness theorem, this means that $\Sigma \models \bot$, and hence $M \models \bot$, which is a contradiction. Hence $\Sigma$ must be consistent.