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Existential quantification forms $\exists x\phi (x,\mathbf{y})$. In a first-order structure, it is true at an assignment to $\mathbf{y}$ when some domain element makes the instance true. In natural deduction, elimination uses a fresh variable for the hypothetical witness; that variable may not escape into the conclusion or the remaining undischarged assumptions.

Existence requires at least one witness, not exactly one. Over the natural numbers $\mathbb{N}=\{0,1,2,\dots \}$, $\exists m(m>5)$ is true: $m=6$ works, as do $7,8,\dots$. In intuitionistic first-order logic, a direct proof supplies a witness together with a proof of the corresponding instance; merely rejecting nonexistence is generally weaker.

Placing existential quantification before universal quantification requires a single witness that works throughout:This first-order sentence is false. For any proposed $m$, choose $n=m$: the required inequality becomes $m>m$. For example, $m=3$ works for $n=0,1,2$ but fails at $n=3$. No finite list of successful checks establishes universal quantification over the natural numbers.

Intuitionistic first-order logic has the usual introduction and elimination rules for logical connectives, quantifiers and logical equality, but no unrestricted double-negation elimination or law of excluded middle. Its proofs retain the witness and logical disjunction information that can be lost in classical first-order logic.

Full intuitionistic Zermelo–Fraenkel set theory uses intuitionistic first-order logic, extensionality, pairing, union, infinity, full separation, power set, collection and epsilon induction. Collection is stronger than replacement intuitionistically. This is distinct from predicative constructive ZF, which restricts separation and replaces full power set. The full theory supports a Boolean-valued inner universe over intuitionistic set theory, giving the relative consistency of classical ZF.

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

A first-order formula $A$ is stable when intuitionistic first-order logic proves $\neg \neg A\to A$. Double negations and logical falsity are stable. Logical conjunction of stable first-order formulas is stable; a logical implication is stable when its consequent is stable; and universal quantification preserves stability. For logical implication, from $\neg \neg (A\to B)$ and $A$, assuming $\neg B$ contradicts the first hypothesis, giving $\neg \neg B$ and then $B$. For universality, obtain $\neg \neg A(t)$ from $\neg \neg \forall xA(x)$ for arbitrary $t$, then use stability and generalize.

Universal quantification forms $\forall x\phi (x,\mathbf{y})$. In a first-order structure, it is true at an assignment to $\mathbf{y}$ when every domain element assigned to $x$ makes the instance true. Its introduction rule requires that the generalized variable not occur freely in undischarged assumptions.

Over the natural numbers, reversing the preceding existential quantification and universal quantification gives a true first-order sentence:Choose $m=n+1$. The computed witnesses for $n=0,1,2,3$ are $m=1,2,3,4$. The choice may depend on the previously given $n$. The formula states that every natural number has a larger natural number; the reversed formula demands one larger than all of them.

The diagram displays a finite window, while the rules $m=n+1$ and $n=m$ settle the two statements for every natural number. This dependence is why quantifier order matters in precise specifications. Choosing one $m$ that works for every $n$ would also provide a witness for each $n$; the reverse inference fails here.

To introduce universal quantification in natural deduction, reason with an arbitrary input, without assuming a special property of it. The construction $m=n+1$ does exactly this and also gives a proof in intuitionistic first-order logic. The introduction and elimination rules for both quantifiers are developed in Frank Pfenning, First-Order Logic and Type Theory, section 4.1.