Metamath Proof Explorer


Theorem oncardid

Description: Any ordinal number is equinumerous to its cardinal number. Unlike cardid , this theorem does not require the Axiom of Choice. (Contributed by NM, 26-Jul-2004)

Ref Expression
Assertion oncardid ⊢ A ∈ On → card ⁡ A ≈ A

Proof

Step Hyp Ref Expression
1 onenon ⊢ A ∈ On → A ∈ dom ⁡ card
2 cardid2 ⊢ A ∈ dom ⁡ card → card ⁡ A ≈ A
3 1 2 syl ⊢ A ∈ On → card ⁡ A ≈ A