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