Metamath Proof Explorer


Theorem kardcard2

Description: Two numerable sets have equal kard cardinalities iff they have equal card cardinalities. This theorem does not depend on the Axiom of Choice. (Contributed by BTernaryTau, 3-Jul-2026)

Ref Expression
Assertion kardcard2 ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ↔ ( card ‘ 𝐴 ) = ( card ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 kardeng ⊢ ( 𝐴 ∈ dom card → ( ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ↔ 𝐴 ≈ 𝐵 ) )
2 1 adantr ⊢ ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ↔ 𝐴 ≈ 𝐵 ) )
3 carden2 ⊢ ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( card ‘ 𝐴 ) = ( card ‘ 𝐵 ) ↔ 𝐴 ≈ 𝐵 ) )
4 2 3 bitr4d ⊢ ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ↔ ( card ‘ 𝐴 ) = ( card ‘ 𝐵 ) ) )