Metamath Proof Explorer


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 e. V -> ( ( kard ` A ) = ( kard ` B ) <-> A ~~ B ) )

Proof

Step Hyp Ref Expression
1 fveqeq2
 |-  ( x = A -> ( ( kard ` x ) = ( kard ` B ) <-> ( kard ` A ) = ( kard ` B ) ) )
2 breq1
 |-  ( x = A -> ( x ~~ B <-> A ~~ B ) )
3 vex
 |-  x e. _V
4 kardval
 |-  ( kard ` x ) = Scott { y | y ~~ x }
5 kardval
 |-  ( kard ` B ) = Scott { y | y ~~ B }
6 3 4 5 karden
 |-  ( ( kard ` x ) = ( kard ` B ) <-> x ~~ B )
7 1 2 6 vtoclbg
 |-  ( A e. V -> ( ( kard ` A ) = ( kard ` B ) <-> A ~~ B ) )