Metamath Proof Explorer


Theorem carden2b

Description: If two sets are equinumerous, then they have equal cardinalities. (This assertion and carden2a are meant to replace carden in ZF without AC.) (Contributed by Mario Carneiro, 9-Jan-2013) (Proof shortened by Mario Carneiro, 27-Apr-2015)

Ref Expression
Assertion carden2b A B card A = card B

Proof

Step Hyp Ref Expression
1 cardne card B card A ¬ card B A
2 ennum A B A dom card B dom card
3 2 biimpa A B A dom card B dom card
4 cardid2 B dom card card B B
5 3 4 syl A B A dom card card B B
6 ensym A B B A
7 6 adantr A B A dom card B A
8 entr card B B B A card B A
9 5 7 8 syl2anc A B A dom card card B A
10 1 9 nsyl3 A B A dom card ¬ card B card A
11 cardon card A On
12 cardon card B On
13 ontri1 card A On card B On card A card B ¬ card B card A
14 11 12 13 mp2an card A card B ¬ card B card A
15 10 14 sylibr A B A dom card card A card B
16 cardne card A card B ¬ card A B
17 cardid2 A dom card card A A
18 id A B A B
19 entr card A A A B card A B
20 17 18 19 syl2anr A B A dom card card A B
21 16 20 nsyl3 A B A dom card ¬ card A card B
22 ontri1 card B On card A On card B card A ¬ card A card B
23 12 11 22 mp2an card B card A ¬ card A card B
24 21 23 sylibr A B A dom card card B card A
25 15 24 eqssd A B A dom card card A = card B
26 ndmfv ¬ A dom card card A =
27 26 adantl A B ¬ A dom card card A =
28 2 notbid A B ¬ A dom card ¬ B dom card
29 28 biimpa A B ¬ A dom card ¬ B dom card
30 ndmfv ¬ B dom card card B =
31 29 30 syl A B ¬ A dom card card B =
32 27 31 eqtr4d A B ¬ A dom card card A = card B
33 25 32 pm2.61dan A B card A = card B