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If a consistent propositional theory locally omits every member of a countable family of propositional types, it has a Boolean valuation omitting them all. One may also impose any finite condition consistent with the theory. Successively extend the condition by a negated member of the next type; local omission preserves consistency. The propositional compactness theorem then supplies the valuation. Countability of the ambient language and decidability of consistency are not needed for this argument.

A partially ordered set is two-dimensional if every one of its finite induced suborders is two-dimensional. Encode two candidate total orders by propositional variables; every finite collection of the order, extension, and intersection clauses concerns a finite induced suborder, so the propositional compactness theorem supplies two global realizing orders.

A propositional type is a set $\Sigma$ of propositional formulas; a Boolean valuation realizes it if every member is true, and omits it if at least one member is false. A consistent propositional theory $T$ locally omits $\Sigma$ if every formula $\theta$ which implies all members of $\Sigma$ modulo $T$ is refutable modulo $T$. Equivalently, whenever $T\cup \{\theta \}$ is consistent, there is a $\sigma \in \Sigma$ for which $T\cup \{\theta ,\neg \sigma \}$ is consistent. For a consistent propositional type, this says that it is a nonprincipal propositional type.

The extended omitting types theorem for propositional logic says that if a consistent $T$ locally omits each of a countable family ${\Sigma }_0,{\Sigma }_1,\dots$, there is one Boolean valuation satisfying $T$ and omitting them all. It also works inside any prescribed finite condition consistent with $T$.

To prove it, start with such a condition ${\theta }_0$, taking $\top$ when none is prescribed. At stage $i$, choose ${\sigma }_i\in {\Sigma }_i$ such thatSuch a choice exists by local omission. More explicitly, failure would imply $T⊢{\theta }_i\to \sigma$ for every $\sigma \in {\Sigma }_i$, hence $T⊢\neg {\theta }_i$, contradicting the stage invariant. Every finite subset of $T\cup \{{\theta }_0\}\cup \{\neg {\sigma }_i:i\in \mathbb{N}\}$ lies inside a consistent stage. By the propositional compactness theorem and the completeness theorem for propositional logic, it has a Boolean valuation. That Boolean valuation satisfies $T$ and makes the selected member ${\sigma }_i$ of every type false. All types are therefore omitted simultaneously. No effective test for consistency is assumed, and the argument does not require the ambient propositional language to be countable. Only the family of omission requirements is countable.

The completeness theorem for propositional logic states that $S\models \varphi$ if and only if $S⊢\varphi$. Soundness follows by checking that axioms are tautologies and inference rules preserve truth. For completeness it suffices to prove that every syntactically consistent set has a model.

The deduction theorem for propositional logic used here is: $S\cup \{\psi \}⊢\chi$ if and only if $S⊢\psi \Rightarrow \chi$, in classical propositional logic. A proof is finite, so the union of a chain of consistent sets remains consistent: a contradiction would already occur in one member. By Zorn's lemma, extend a consistent $S$ to a maximal consistent set $T$. If $\psi \notin T$, adding $\psi$ yields a contradiction; the deduction theorem gives $T⊢\neg \psi$, and maximal consistency puts $\neg \psi$ in $T$. Conversely $\psi$ and $\neg \psi$ cannot both belong to $T$. Maximal consistency also makes $T$ deductively closed.

Assign each primitive proposition the truth value true precisely when it belongs to $T$. The maximal consistent set truth lemma follows by induction on formulas. For negation use the preceding dichotomy. For implication, if $\neg \psi \in T$ or $\chi \in T$, the classical implication axioms put $\psi \Rightarrow \chi$ in $T$; if $\psi \in T$ and $\neg \chi \in T$, modus ponens prevents that implication from belonging to $T$. This is exactly its truth table. Other connectives are defined from these or handled by the same induction. The resulting valuation satisfies all of $T$, hence all of $S$.

If $S⊬\varphi$, the deduction theorem and classical contradiction rules show $S\cup \{\neg \varphi \}$ is consistent. Its model falsifies $\varphi$, so $S/\models \varphi$. This proves completeness.

The propositional compactness theorem says $S$ is satisfiable if every finite subset is satisfiable. Otherwise completeness would make $S$ inconsistent, and a finite proof of contradiction would involve only finitely many assumptions, contradicting their satisfiability. Equivalently, every semantic consequence of $S$ is a consequence of a finite subset. The decidability theorem for propositional logic says there is an algorithm deciding whether a formula is provable with no assumptions: enumerate all truth assignments to its finitely many primitive propositions, check whether it is a tautology, and use soundness and completeness to identify tautologies with theorems. This does not claim an algorithm for arbitrary infinitely presented theories.

Finally, the valuation assigning false to every $p_n$ satisfies all the implications in $S$ and falsifies $p_1$. By soundness, $S⊬p_1$.

The completeness theorem for propositional logic, together with soundness, states $S⊢\phi$ if and only if $S\models \phi$. In particular, an inconsistent set of formulae has a finite formal proof of contradiction. Such a proof uses only finitely many assumptions, yielding the propositional compactness theorem: if every finite subset of $S$ is satisfiable, then $S$ is satisfiable.

The decidability theorem for propositional logic concerns theoremhood of a formula, or entailment from a finite list of premises. Formulae involve finitely many variables, so their valuations can be tested by a finite truth table; completeness identifies semantic validity with provability. Equivalently, dovetail effective proof enumeration with the finite search for a countervaluation: completeness ensures that one succeeds. This is not a decidability claim for consequences of an arbitrary noncomputable infinite theory.

An infinite collection of distinct tautologies is a finitary set of formulae, since its deductive closure equals that of the empty set. By contrast, the set of independent variables $\{p_i:i\ge 1\}$ is not finitary. If a finite set $T$ had the same closure, it would mention only finitely many variables. It is satisfiable because it is a consequence of the all-true model of the $p_i$; changing an unmentioned $p_j$ to false preserves $T$ and contradicts $T\models p_j$.

Under the final hypothesis, $A\cup B$ has no model. The propositional compactness theorem gives finite $A^{\prime }\subset A$, $B^{\prime }\subset B$ with $A^{\prime }\cup B^{\prime }$ inconsistent. Any valuation satisfying $A^{\prime }$ cannot satisfy $B$, because it would then satisfy $B^{\prime }$; by the exactly-one hypothesis it must satisfy $A$. The reverse implication holds by inclusion. Thus $A$ and $A^{\prime }$ have identical models, and similarly $B$ and $B^{\prime }$. Completeness gives identical deductive closures, proving

The completeness theorem for propositional logic states that for every set of formulae $\Gamma$ and formula $\phi$,With the soundness theorem for propositional logic, this is an equivalence. It is also equivalent to saying that every consistent set of formulae has a model.

The usual countable proof lists all formulae and extends a consistent theory one formula at a time. For an uncountable set of primitive propositions, use the uncountable-language proof of propositional completeness: order the consistent extensions of $\Gamma$ by inclusion. The union of every chain is consistent, since a proof of a contradiction would use only finitely many assumptions and hence would already lie in one member of the chain. Zorn lemma gives a maximal consistent set in propositional logic $M$. Define a Boolean valuation by $v(p)=1$ exactly when $p\in M$. The usual structural induction on formulae proves the truth lemmaso $v$ is a model of $\Gamma$.

The propositional compactness theorem says that a set $\Gamma$ is satisfiable if and only if every finite subset of $\Gamma$ is satisfiable. Only the forward implication is immediate. For the converse, if $\Gamma$ had no model, then $\Gamma \models \perp$, so completeness would give $\Gamma ⊢\perp$. Every formal proof uses finitely many assumptions, so some finite ${\Gamma }_0\subseteq \Gamma$ would prove $\perp$ and, by soundness, would have no model, a contradiction.

The decidability theorem for propositional logic says that there is an algorithm deciding whether a propositional formula is a theorem. A formula contains only finitely many primitive propositions, even when the full language is uncountable. Its finite truth table decides whether it is valid, and completeness says that validity is equivalent to theoremhood.

Now let $(X,<)$ be the given partially ordered set. For every distinct $x,y\in X$, introduce propositions $p_{xy}$ and $q_{xy}$. Form a propositional theory $T$ containing the following clauses:

The first two families say that $p$ and $q$ encode strict total orders. The third makes both orders extend $<$, while the fourth makes them disagree on every incomparable pair.

Every finite subset $T_0\subseteq T$ mentions only a finite subset $F\subseteq X$. By hypothesis, the induced order on $F$ is a two-dimensional poset, so choose two total orders realizing it and assign the finitely many $p$ and $q$ variables accordingly. This satisfies $T_0$. Thus every finite subset of $T$ is satisfiable, and propositional compactness gives a model of all of $T$.

Define $x{<}_{1}y$ when $p_{xy}$ is true and $x{<}_{2}y$ when $q_{xy}$ is true. The clauses make ${<}_1$ and ${<}_2$ total orders on $X$. They both contain the original order, and their intersection contains no incomparable pair. Hencewhich proves the finite-local characterization of two-dimensional partially ordered sets.

Introduce a propositional variable $P_{x,i}$ for every $x\in X$ and $i\in \{1,\dots ,100\}$. Form a theory containing, for every $x$, clauses saying that exactly one ofis true, and, for every $y\in L_x$ and every $i$, the clauseEvery finite subset of these clauses mentions only finitely many elements, forming a finite set $Y\subset X$. The assumed coloring $f_Y$ gives that finite subset a model. The propositional compactness theorem, equivalently the propositional consequence of first-order compactness, therefore gives a model of all the clauses.

For each $x$, let $F(x)$ be the unique $i$ for which $P_{x,i}$ is true. The edge clauses ensure that $F(x)\ne F(y)$ whenever $y\in L_x$. Hencehas the required property.

No such set of propositions exists. Suppose that $S$ had precisely the total injections as its models, fix $x_0\in X$, and adjointo $S$. Every finite subset of the enlarged theory has a model. Indeed, it excludes only finitely many possible values $F\subset Y$ for $x_0$. Starting with any injection $h:X\to Y$, either $h(x_0)\notin F$, or one may choose $x_1\ne x_0$ with $h(x_1)\notin F$ and swap the values of $x_0$ and $x_1$.

The propositional compactness theorem would therefore give a model of the whole enlarged theory. Its relation $f_v$ is a model of $S$ but has no value at $x_0$, contradicting the assumed description of the models of $S$. This is the compactness obstruction to expressing totality over an infinite codomain.

The soundness theorem for propositional logic says that if $\Gamma ⊢t$, then every valuation satisfying every member of $\Gamma$ also satisfies $t$.

The proposed function need not respect the connectives. For example, if a primitive proposition $p$ is independent of $S$, then neither $p$ nor $\neg p$ is provable from $S$, so the definition gives

Order the consistent supersets of $S$ by inclusion. The union of a chain is consistent, since a finite proof of a contradiction would already use assumptions from one member of the chain. Zorn lemma therefore gives a maximal consistent extension $T$. It is deductively closed: if $T⊢t$, adjoining $t$ preserves consistency, so maximality forces $t\in T$. Moreover, for every $t$, exactly one of $t$ and $\neg t$ belongs to $T$. They cannot both belong by consistency; if neither belonged, the inconsistency of both proper extensions would give $T⊢\neg t$ and $T⊢\neg \neg t$, again a contradiction. The usual induction on formulae now shows thatdefines a valuation satisfying $T$, and hence $S$.

Now suppose every finite subset of $S$ has a model. By soundness every finite subset is consistent. Any proof of a contradiction from $S$ uses only finitely many assumptions, so $S$ itself is consistent. Applying the preceding maximal-consistent-extension construction gives a model of $S$. This proves the propositional compactness theorem.