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An ordinal $\kappa$ is a cardinal number precisely when no smaller ordinal has a bijection onto it. Ordinalhood and the graph predicate for a bijection are bounded formulas in set theory, so $\mathrm{Ord}(\kappa )\wedge \forall f\forall \alpha \in \kappa \neg \mathrm{Bij}(f,\alpha ,\kappa )$ is a Pi-one formula modulo ZF. This does not assert that every set can be assigned an ordinal cardinal without the axiom of choice.

Let $\mathrm{Map}(f,\delta ,\kappa )$ be the bounded graph predicate for a function $\delta \to \kappa$. For an infinite cardinal number, regularity is expressed byThe predicate $\mathrm{InfCard}$ is the cardinal predicate in (i), conjoined with the bounded condition that $\kappa$ is a nonzero limit ordinal. For an infinite cardinal number, this latter condition holds automatically; it excludes the finite cardinals. The expression $f(\xi )<\beta$ abbreviates bounded graph membership with the output lying in $\beta$.

Every quantifier after $\forall f$ is bounded, and the cardinal predicate is already ${\Pi }_1$. Their conjunction can be placed in universal form, so this is a Pi-one formula modulo ZF. Its mathematical content is that no function from a smaller ordinal is cofinal in $\kappa$. By the definition of cofinality, that is equivalent to $\mathrm{cf}(\kappa )=\kappa$. Hence regular cardinalhood is Pi-one definable. This uses the standard convention that a regular cardinal is infinite.

Use the Lévy hierarchy, with bounded formulas in set theory as the quantifier-free level for this purpose. Let $\mathrm{Ord}(\kappa )$ be the bounded ordinal predicate, and let $\mathrm{Bij}(f,\alpha ,\kappa )$ be the bounded predicate saying that $f$ is a graph of a bijection from $\alpha$ onto $\kappa$. Then cardinalhood is expressed byAll quantifiers in $\mathrm{Ord}$ and $\mathrm{Bij}$ are bounded by the displayed sets and the graph. Ordered-pair membership, graph functionality, injectivity and surjectivity have bounded expansions using the usual set coding of ordered pairs. The only unbounded quantifier shown is universal over $f$. The formula is therefore ${\Pi }_1$ and in particular a Pi-one formula modulo ZF.

In ZF, a cardinal number is an ordinal not in bijection with any smaller ordinal, so this formula has exactly the required meaning. Thus cardinalhood is Pi-one definable; the argument does not use the axiom of choice.

A syntactically ${\Pi }_1$ first-order formula has an unbounded universal quantifier block followed by a bounded formula in set theory. Universal quantification makes it downward absolute between transitive models. A Pi-one formula modulo ZF may have another syntax but has such an equivalent provable in ZF.

An infinite cardinal number $\kappa$ is a regular cardinal if every function from an ordinal $\delta <\kappa$ into $\kappa$ has bounded range. Write this as $\forall f\forall \delta \in \kappa (\mathrm{Map}(f,\delta ,\kappa )\Rightarrow \exists \beta \in \kappa \forall \xi \in \delta f(\xi )<\beta )$. The graph predicates are bounded formulas in set theory, and cardinalhood is Pi-one definable, so adding infinite cardinalhood yields a Pi-one formula modulo ZF.