Metamath Proof Explorer


Theorem kardcard2a

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

Ref Expression
Assertion kardcard2a
|- ( ( ( card ` A ) = ( card ` B ) /\ ( card ` A ) =/= (/) ) -> ( kard ` A ) = ( kard ` B ) )

Proof

Step Hyp Ref Expression
1 carden2a
 |-  ( ( ( card ` A ) = ( card ` B ) /\ ( card ` A ) =/= (/) ) -> A ~~ B )
2 fvfundmfvn0
 |-  ( ( card ` A ) =/= (/) -> ( A e. dom card /\ Fun ( card |` { A } ) ) )
3 2 simpld
 |-  ( ( card ` A ) =/= (/) -> A e. dom card )
4 kardeng
 |-  ( A e. dom card -> ( ( kard ` A ) = ( kard ` B ) <-> A ~~ B ) )
5 3 4 syl
 |-  ( ( card ` A ) =/= (/) -> ( ( kard ` A ) = ( kard ` B ) <-> A ~~ B ) )
6 5 adantl
 |-  ( ( ( card ` A ) = ( card ` B ) /\ ( card ` A ) =/= (/) ) -> ( ( kard ` A ) = ( kard ` B ) <-> A ~~ B ) )
7 1 6 mpbird
 |-  ( ( ( card ` A ) = ( card ` B ) /\ ( card ` A ) =/= (/) ) -> ( kard ` A ) = ( kard ` B ) )