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Prove without using compactness that if $T$ is a maximal $L$ theory such that every finite subset of $T$ is consistent, then either $\sigma \in T$ or $\lnot \sigma \in T$. Here, by maximal, we mean maximal with respect to this property.

Attempt:

Let $T$ be an $L$-theory with the given property and fix an $L$ sentence $\sigma$. Suppose both $\sigma, \lnot \sigma \in T$. Then $S = \{\sigma, \lnot \sigma \} \subseteq T$ is a finite subset of $T$, so $S$ is consistent. Thus there exists a model of $S$, say $\mathcal{S}$ such that $\mathcal{S}$ satisfies both $\sigma$ and $\lnot \sigma$. This is impossible, so we conclude that for any $L$ sentence $\sigma$, at most one of $\sigma, \lnot \sigma \in T$.

Suppose $\sigma, \lnot \sigma$ are both not in $T$. Then consider $T_1 = T \cup \{\sigma\}$ and $T_2 = T \cup \{\lnot \sigma\}$. $T$ is contained in $T_1$ and $T_2$ and it is maximal, so it must be that $T_1, T_2$ each contain finite subsets which are not consistent. These subsets must contain $\sigma, \lnot \sigma$ respectively because otherwise they would be finite subsets of $T$, which are consistent by assumption. Thus we can write them as $T' \cup \{\sigma\}, T'' \cup \{\lnot \sigma\}$ resp.

I am not sure how to proceed from here. Any hints would be great!

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  • $\begingroup$ First part: Suppose both σ,¬σ∈T. Then S={σ,¬σ}⊆T is a finite subset of T tha is inconsistent. Contradiction. $\endgroup$ Nov 19, 2020 at 15:13
  • $\begingroup$ Second part: the two are supersets of T: thus T is not maximal. But... you have to prove that at least one of them is consistent. $\endgroup$ Nov 19, 2020 at 15:15
  • $\begingroup$ If $T'\cup\{\sigma\}$ and $T''\cup\{\lnot\sigma\}$ are finite, inconsistent theories, then $T'\cup T''\subseteq T$ is a finite, inconsistent theory. $\endgroup$ Nov 19, 2020 at 15:23
  • $\begingroup$ Why is that theory inconsistent @spaceisdarkgreen? $\endgroup$
    – John
    Nov 19, 2020 at 15:26
  • $\begingroup$ @John Thought you were asking for hints! $\endgroup$ Nov 19, 2020 at 15:27

1 Answer 1

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For the second part, we will prove that $T' \cup T''$ is a finite inconsistent subset of $T$. This contradicts $T$ having only finite consistent subsets, so we will be done.

$T′∪T′′∪\{σ\}$ is inconsistent because it is a superset of $T′∪\{σ\}$ and $T′∪T′′∪\{¬σ\}$ is inconsistent because it is a superset of $T′′∪\{¬σ\}$. Thus $T′∪T′′$ must be inconsistent, otherwise if a model existed, then it would satisfy either $σ$ or $¬σ$, so one of $T′∪T′′∪\{σ\}, T′∪T′′∪\{¬σ\}$ would be consistent.

Note : Supersets of inconsistent theories are inconsistent. If a model exists for $T$, a theory which contains $Q$, then this model satisfies all the $L$ sentences in $T$, so it also satisfies all the $L$ sentences in $Q$. Thus it must be a valid model for $Q$ as well, so $Q$ must be consistent too.

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