A theory $T$ of $L$-sentences is said to be consistent if there exists a model of $T$. More precisely, there exists an $L$-structure $\mathcal{M}$ such that $M$ satisfies $\sigma$ for each $\sigma \in T$.
How does this definition of consistent compare with the more 'intuitive' definition of consistent, that is a theory $T$ is said to be consistent if and only if for all $L$-sentences $\sigma$, $T$ does not contain both $\sigma$ and $\lnot \sigma$. I think they should be equivalent but I am unable to come up with a proof.
I thought that if $T$ contains both, then any model of $T$ will satisfy both, so there cannot exist a model of $T$. But I am not sure how to show that there cannot exist a model of $T$.
Also, if $T$ is a consistent theory, then it has a model $\mathcal{M}$ and if there exists $\sigma, \lnot \sigma \in T$, then $\mathcal{M}$ satisfies both $\sigma$ and $\lnot \sigma$, which means $\mathcal{M}$ is not a valid $L$-structure. Again, I am not sure how to show that it is not a valid $L$-structure from the definitions..
Assume that $\sigma,\lnot \sigma \in T$.
A model of $T$ is a structure $\mathcal M$ such that:
Thus: $\mathcal M \vDash \sigma$ and $\mathcal M \vDash \lnot \sigma$.
But one of the clause of the definition of the satisfaction relation is:
The usual terminology is: a theory $T$ is satisfiable iff it has a model.
Thus, the above argument is: if $T$ is inconsistent, then it is unsatisfiable, i.e. if $T$ is satisfiable, then is consistent. [Thanks to David.]
The other part of the equivalence: $T$ is satisfiable iff it is consistent, is the so-called Model existence lemma, a more general version of Gödel's completeness theorem.