Let contain a binary relation symbol , and let interpret as the usual strict order. For an assignment , consider
- The quantifiers vary the assignment in turn:
- Every $n$ has a larger natural number, so $\hat b(\varphi_1)=1$. By contrast, $\hat b(\varphi_2)=0$: no $n$ is less than every natural number. The order of the quantifiers changes the claim.
- Write $\mathcal M\models\varphi$ if $\mathcal M\models\varphi[b]$ for every assignment $b$.
- **[[Knowledge/Coincidence Lemma (First-Order Logic)|Coincidence lemma]]:** If $b(x)=b'(x)$ for every variable $x$ free in $\varphi$, then $\hat b(\varphi)=\hat b'(\varphi)$. The values of variables that are not free in $\varphi$ do not affect its truth value.
- In $\exists y\,xRy$, the occurrence of $x$ is free and the occurrence of $y$ is [[Knowledge/Bound Variable (First-Order Logic)|bound]]. In $(\exists x\,xRy)\lor xRy$, the first occurrence of $x$ is bound and the second is free; $y$ is free in both disjuncts.
- $x$ is free in $\forall z\,\psi$ exactly when $x\neq z$ and $x$ is free in $\psi$. The same applies to $\exists z$.
- A [[Knowledge/Sentence (First-Order Logic)|sentence]] is a formula with no free variables. Its truth in a structure does not depend on the assignment. A sentence need not be a theorem.
- For example, $(\mathbb N,<)\models\forall x\,\exists y\,xRy$, whereas $(\mathbb N,>)\not\models\forall x\,\exists y\,xRy$: with $R$ interpreted as $>$, the choice $x=0$ has no witness.
where $\mathcal M$ ranges over $\mathcal L$-structures. Thus entailment concerns all models of the premises, not just one structure.
Arithmetic axioms
In the language with , successor , addition and multiplication, consider the following axioms; the variables in the last four equations are universally quantified:
- These equations describe zero, successor and the operations; the [[Knowledge/Peano Axioms|Peano axioms]] also include induction. The displayed list alone does not state the induction schema.
Set-theoretic language
With membership as the only non-logical symbol, expressions such as are shorthand, not terms of the language. For example,
- The same idea applies to nested singletons: $x=\{\{y\}\}$ can be expressed by introducing an object for $\{y\}$ and describing its members.
Singleton and weak singleton statements
Let state that every object has a singleton, and let state only that every object belongs to something:
- $\theta_1\models\theta'_1$: a singleton containing $x$ is a witness for the weaker statement. Equivalently, $\models\theta_1\to\theta'_1$.
- The converse fails. Interpret $\in$ as $<$ in $\mathcal N=(\mathbb N,<)$. Then $\mathcal N\models\theta'_1$, since $n<n+1$ for every $n$. But $\mathcal N\not\models\theta_1$: no $k$ has exactly $2$ as its only predecessor. Indeed, if $2<k$, then $0<k$ as well. Hence $\theta'_1\not\models\theta_1$.
- The quantifier $\forall z$ requires the equivalence to hold for every object, so $y$ has exactly $x$ as a member.
Further set-existence statements
The pairing statement says that for any there is an object whose members are exactly and :
- Existence of an empty set is expressed by $\exists x\,\forall y\,(y\notin x)$.
- For the power-set statement, $z\subseteq x$ abbreviates $\forall u\,(u\in z\to u\in x)$:
- In $(\mathbb N,<)$, with $<$ interpreting membership, the object $4$ acts as a power set of $3$ under this formula: $z\subseteq 3$ holds precisely for $z\leq 3$, and these are precisely the predecessors of $4$. This is a statement about the interpretation of the formula, not the usual power set of the number $3$.
- The union statement says that every $A$ has a union $B$: