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 ) ) |
| 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 ) ) |