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A combinator is a lambda term with no free variables.

The logic $L_{{\omega }_1,\omega }$ extends first-order logic by countable logical conjunctions and logical disjunctions. Each formula has finitely many free variables, and quantifier strings are finite. In a countable first-order language, every countable structure has a Scott sentence in this logic that characterizes its isomorphism class among countable structures.

Every countable structure in a countable first-order language has a Scott sentence whose countable models are precisely its isomorphic copies. In particular two countable structures with the same countable infinitary logic sentences are isomorphic. Countability of both structures is essential to the back-and-forth conclusion; the sentence need not exclude uncountable models.

Let $M$ be a nonempty countable structure in a countable first-order language. The countable infinitary logic $L_{{\omega }_1,\omega }$ permits countable logical conjunctions and logical disjunctions, but only finite strings of quantifiers and finitely many free variables in each formula. We construct a Scott formula ${\varphi }_{\bar{a}}^{\alpha }(\bar{x})$ for every finite tuple $\bar{a}$ from $M$ and every countable ordinal $\alpha$.

At stage zero, let ${\varphi }_{\bar{a}}^0$ be the conjunction of all atomic formulas true of $\bar{a}$ and the negations of all atomic formulas false of $\bar{a}$. Include atomic formulas involving arbitrary terms and constants, not just relation symbols applied directly to variables. There are only countably many such formulas. This complete atomic description ensures that matching tuples determine a partial isomorphism of structures.

At successors putAt a nonzero limit ordinal $\lambda <{\omega }_1$, put ${\varphi }_{\bar{a}}^{\lambda }={\bigwedge }_{\beta <\lambda }{\varphi }_{\bar{a}}^{\beta }$. All these are formulas of countable infinitary logic: each indexing set is countable, and the free variables are only the fixed finite tuple. Rename bound variables when necessary. Transfinite induction also shows $M\models {\varphi }_{\bar{a}}^{\alpha }(\bar{a})$.

For tuples of the same length within $M$, define $\bar{a}{\equiv }_{\alpha }\bar{b}$ by $M\models {\varphi }_{\bar{a}}^{\alpha }(\bar{b})$. Induction identifies this with the usual symmetric back-and-forth method equivalence: the tuples have the same atomic description at stage zero, and at a successor every one-element extension on either side has a matching extension at the previous stage. Thus these are decreasing equivalence relations, simultaneously for every finite tuple length.

There are only countably many pairs of finite tuples in $M$. A pair can cease to be equivalent at most once. The supremum of the first separation stages of all pairs that separate below ${\omega }_1$ is a countable ordinal. Choose a countable $\alpha$ at least that supremum. No pair can first separate at $\alpha +1$, soThis justifies stabilization without assuming that all tuples stabilize at one predetermined finite stage.

Now form the following Scott sentence, where the case $n=0$ has no displayed variables or quantifiers:It is a sentence of countable infinitary logic, because the family of all finite tuples is countable. The stabilization above and the truth of ${\varphi }_{\varnothing }^{\alpha }$ show $M\models {\sigma }_M$.

Suppose a countable structure $N$ satisfies ${\sigma }_M$. Start with the empty matching tuples, and maintain $N\models {\varphi }_{\bar{a}}^{\alpha }(\bar{b})$. The corresponding conjunct of ${\sigma }_M$ upgrades this to $N\models {\varphi }_{\bar{a}}^{\alpha +1}(\bar{b})$. The existential conjuncts extend the match by any specified element of $M$. The universal-disjunction conjunct extends it by any specified element of $N$. The atomic formula information makes a new element on one side match a new element on the other, and makes repeated elements agree with their earlier matches.

Enumerate both structures and alternate these two extension steps, including the least element not yet covered at each step. The union is a bijection preserving and reflecting every atomic formula, hence an isomorphism; for function symbols, eventually include both a tuple and the value of its function term to see explicitly that the function is preserved. For finite structures the same construction stops when both are covered. Conversely every isomorphic copy of $M$ satisfies ${\sigma }_M$, since isomorphisms preserve formulas of countable infinitary logic by induction on their construction. ThereforeIf countable $M,N$ have the same $L_{{\omega }_1,\omega }$ sentences, $N$ satisfies this Scott sentence of $M$ and is isomorphic to $M$. This proves Scott isomorphism theorem.

Use the Gödel-Gentzen negative translation, writing $A^N$ for the translated first-order formula. On atomic formulas $P$, including logical equality, put $P^N=\neg \neg P$, and put ${\perp }^N=\perp$. Extend recursively byLogical negation abbreviates logical implication to logical falsity, so $(\neg A{)}^N=\neg A^N$. The displayed logical disjunction and existential clauses are intuitionistically equivalent to the usual negative clauses $\neg (\neg A^N\wedge \neg B^N)$ and $\neg \forall x\neg A^N$, respectively. Thus these clauses specify the same negative interpretation. Translation commutes with capture-avoiding substitution of terms for free variables.

First prove by structural induction the stability of a formula under double negation:Every double logical negation is stable, since intuitionistic first-order logic proves $\neg \neg \neg \neg C\to \neg \neg C$; logical falsity is stable as well. Stability passes to logical conjunction by obtaining double logical negation of each component. For a logical implication $C\to D$ with stable $D$, assume $\neg \neg (C\to D)$ and $C$. An assumption $\neg D$ would give $\neg (C\to D)$, a contradiction, so $\neg \neg D$ and then $D$. For a universal first-order formula, $\neg \neg \forall xC(x)$ implies $\neg \neg C(y)$ for arbitrary fresh $y$; use stability pointwise and generalize. Logical disjunction and existential translations are already double negations. This proves all cases.

Now regard classical first-order natural deduction as intuitionistic natural deduction plus unrestricted double-negation elimination, and induct on a classical derivation. The rules for logical conjunction, logical implication, universal quantification and logical falsity translate directly. Logical disjunction introduction gives $A^N\vee B^N$ and then its double logical negation. For logical disjunction elimination, the translated premise is $\neg \neg (A^N\vee B^N)$ and the two translated branches yield the stable conclusion $C^N$. Assuming $\neg C^N$ makes each branch contradictory, giving $\neg A^N$ and $\neg B^N$, hence $\neg (A^N\vee B^N)$, which contradicts the premise. We have $\neg \neg C^N$ and remove it by stability.

Existential introduction likewise adds a double logical negation to the ordinary witness introduction. In existential elimination, the witness branch $A^N(y)⊢C^N$ has the original fresh-variable condition. Assuming $\neg C^N$ makes that branch give $\neg A^N(y)$; generalization gives $\forall y\neg A^N(y)$ and therefore $\neg \exists y{A}^N(y)$, contradicting the translated premise. Again stability supplies $C^N$. The universal-rule side conditions are preserved because the translation introduces no new free variables.

For logical equality, reflexivity gives $t=t$ and then its double logical negation. For substitution, from $\neg \neg (s=t)$ and $A^N(s)$, temporarily assume $\neg A^N(t)$. An assumption $s=t$ would transport $A^N(s)$ to $A^N(t)$ by intuitionistic logical equality substitution, so it gives a contradiction and hence $\neg (s=t)$. The first premise contradicts this. Thus $\neg \neg A^N(t)$ holds and stability gives $A^N(t)$. Finally a classical double-negation elimination step translates precisely to $\neg \neg A^N\to A^N$, already proved. Every rule is therefore covered, giving negative translation of a classical proof:In particular a classical thesis has an intuitionistic, constructive translated proof. If nonlogical assumptions are present, they must be translated too.

In classical logic, double-negation elimination gives $P^N\leftrightarrow P$ on atoms. Structural induction propagates equivalence through logical conjunction, logical implication and both quantifiers, and removes the added double negations on logical disjunction and existence. Therefore