Metamath Proof Explorer


Theorem kardcard

Description: Two sets have equal kard cardinalities iff they have equal card cardinalities. This theorem depends on the Axiom of Choice. (Contributed by BTernaryTau, 3-Jul-2026)

Ref Expression
Assertion kardcard ( ( 𝐴𝑉𝐵𝑊 ) → ( ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ↔ ( card ‘ 𝐴 ) = ( card ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 axac3 CHOICE
2 ackardcard ( CHOICE → ( ( 𝐴𝑉𝐵𝑊 ) → ( ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ↔ ( card ‘ 𝐴 ) = ( card ‘ 𝐵 ) ) ) )
3 1 2 ax-mp ( ( 𝐴𝑉𝐵𝑊 ) → ( ( kard ‘ 𝐴 ) = ( kard ‘ 𝐵 ) ↔ ( card ‘ 𝐴 ) = ( card ‘ 𝐵 ) ) )