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A logical connective forms a compound proposition from component propositions; logical truth and logical falsity are the zero-input cases. Logical negation has one input, while logical conjunction, logical disjunction, and logical implication each have two. In classical propositional logic, these logical connectives are truth-functional: their output value is determined by their input values, as listed in a truth table. This does not characterize every kind of logical connective in every logic. See forall x: Calgary, section 10.1.

Under the Curry-Howard correspondence, Intuitionistic propositional logic propositions are types and proofs are typed terms. Assumptions correspond to typed variables. The logical implication $A\to B$ corresponds to the function type, logical conjunction $A\wedge B$ to the product type, logical disjunction $A\vee B$ to the sum type, logical truth to the unit type, and logical falsity to the empty type.

The rules of natural deduction become term constructors. Implication introduction sends a derivation of $B$ from a variable $x:A$ to the abstraction $\lambda x.M:A\to B$, while implication elimination becomes application $MN$. Pairing and projection implement conjunction, and injections with case analysis implement disjunction. Under this correspondence, normalization of proofs is computation by beta reduction in the simply typed lambda calculus.

Let $\vec{a}\in A^n$, with $A$ finite, and put $T=\mathrm{Th}(\mathcal{A})$. There are only finitely many competing tuples $\vec{b}\in A^n$. For each whose complete type differs from that of $\vec{a}$, choose a logical formula ${\varphi }_{\vec{b}}$ true at $\vec{a}$ and false at $\vec{b}$. Such a logical formula exists by the inequality of their complete types, negating a distinguishing logical formula if necessary.

Let $\theta$ be the conjunction of this finite list, or logical truth if the list is empty. It is true at $\vec{a}$, and every tuple satisfying it has the same partial type as $\vec{a}$. For each $\psi \in \mathrm{tp}(\vec{a})$, the first-order sentenceis therefore true in $\mathcal{A}$ and belongs to $T$. The logical formula $\theta$ is first-order consistent with $T$ because it is realized at $\vec{a}$. It is consequently a principal formula of a type of the partial type, proving every partial type realized in a finite first-order structure is principal. Finiteness of the first-order structure, rather than finiteness of the first-order language, is what makes the conjunction finite.

Add a predicate symbol $P_{\theta }$ for each $L$-formula $\theta (\vec{x})$, of arity equal to its displayed free-variable tuple. Let $T^{+}$ consist of $T$ together with all defining axiomsEvery first-order model of $T$ has a unique such expansion, interpreting the new predicate symbols by their defining sets. Conversely the reduct of every first-order model of $T^{+}$ satisfies $T$.

Given an arbitrary $L^{+}$-formula $\psi (\vec{x})$, replace each new atomic predicate symbol by its original $L$-definition, renaming bound variables to avoid capture. This produces an $L$-formula ${\psi }^{-}(\vec{x})$ withThe defining axiom for that particular logical formula now gives an atomic replacement:Thus the expanded first-order theory eliminates quantifiers. For a first-order sentence, the same translation is an $L$-sentence, decided by completeness of $T$; hence every expanded first-order sentence is decided and $T^{+}$ is complete. Empty free-variable tuples can be handled by zero-ary predicate symbols, or by the usual logical truth and falsity constants. This definitional expansion is called Morleyization.

For each $0\le k\le 2^n$, a propositional formula over $n$ specified variables can have exactly $k$ satisfying assignments and length $O(n^2)$. For $0<k<2^n$, interpret a Boolean valuation as the binary number $x_1\cdots x_n$ and test whether it is less than $k=b_1\cdots b_n$.

For every position $i$ with $b_i=1$, use a conjunction asserting that all earlier bits equal $b_j$ and $x_i=0$, leaving later bits unrestricted. Take the disjunction of these terms. Each satisfying assignment has a unique first differing position, so the terms are disjoint, and their counts sum to ${\sum }_{i:b_i=1}{2}^{n-i}=k$. Their combined length is $O(1+2+\cdots +n)=O(n^2)$. Use logical falsity and logical truth at the two endpoints. The count is over all $n$ variables, including those absent from an individual term or from the whole formula.