Theorem kardeng

Description: Two sets are equinumerous iff their kard cardinal numbers are equal. Unlike carden , this theorem does not depend on the Axiom of Choice, but it does depend on the Axiom of Regularity and the Axiom of Infinity. (Contributed by BTernaryTau, 3-Jul-2026)

	Ref	Expression
Assertion	kardeng	⊢ ( 𝐴 ∈ 𝑉 → ( ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ↔ 𝐴 ≈ 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		fveqeq2	⊢ ( 𝑧 = 𝐴 → ( ( kard ‘ 𝑧 ) = ( kard ‘ 𝐵 ) ↔ ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ) )
2		breq1	⊢ ( 𝑧 = 𝐴 → ( 𝑧 ≈ 𝐵 ↔ 𝐴 ≈ 𝐵 ) )
3		vex	⊢ 𝑧 ∈ V
4		kardval2	⊢ ( kard ‘ 𝑧 ) = { 𝑥 ∣ ( 𝑥 ≈ 𝑧 ∧ ∀ 𝑦 ( 𝑦 ≈ 𝑧 → ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) ) ) }
5		kardval2	⊢ ( kard ‘ 𝐵 ) = { 𝑥 ∣ ( 𝑥 ≈ 𝐵 ∧ ∀ 𝑦 ( 𝑦 ≈ 𝐵 → ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) ) ) }
6	3 4 5	karden	⊢ ( ( kard ‘ 𝑧 ) = ( kard ‘ 𝐵 ) ↔ 𝑧 ≈ 𝐵 )
7	1 2 6	vtoclbg	⊢ ( 𝐴 ∈ 𝑉 → ( ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ↔ 𝐴 ≈ 𝐵 ) )