Metamath Proof Explorer


Theorem carddom2

Description: Two numerable sets have the dominance relationship iff their cardinalities have the subset relationship. See also carddom , which uses AC. (Contributed by Mario Carneiro, 11-Jan-2013) (Revised by Mario Carneiro, 29-Apr-2015)

Ref Expression
Assertion carddom2 ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) ↔ 𝐴𝐵 ) )

Proof

Step Hyp Ref Expression
1 carddomi2 ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) → 𝐴𝐵 ) )
2 brdom2 ( 𝐴𝐵 ↔ ( 𝐴𝐵𝐴𝐵 ) )
3 cardon ( card ‘ 𝐴 ) ∈ On
4 3 onelssi ( ( card ‘ 𝐵 ) ∈ ( card ‘ 𝐴 ) → ( card ‘ 𝐵 ) ⊆ ( card ‘ 𝐴 ) )
5 carddomi2 ( ( 𝐵 ∈ dom card ∧ 𝐴 ∈ dom card ) → ( ( card ‘ 𝐵 ) ⊆ ( card ‘ 𝐴 ) → 𝐵𝐴 ) )
6 5 ancoms ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( card ‘ 𝐵 ) ⊆ ( card ‘ 𝐴 ) → 𝐵𝐴 ) )
7 domnsym ( 𝐵𝐴 → ¬ 𝐴𝐵 )
8 4 6 7 syl56 ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( card ‘ 𝐵 ) ∈ ( card ‘ 𝐴 ) → ¬ 𝐴𝐵 ) )
9 8 con2d ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( 𝐴𝐵 → ¬ ( card ‘ 𝐵 ) ∈ ( card ‘ 𝐴 ) ) )
10 cardon ( card ‘ 𝐵 ) ∈ On
11 ontri1 ( ( ( card ‘ 𝐴 ) ∈ On ∧ ( card ‘ 𝐵 ) ∈ On ) → ( ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) ↔ ¬ ( card ‘ 𝐵 ) ∈ ( card ‘ 𝐴 ) ) )
12 3 10 11 mp2an ( ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) ↔ ¬ ( card ‘ 𝐵 ) ∈ ( card ‘ 𝐴 ) )
13 9 12 syl6ibr ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( 𝐴𝐵 → ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) ) )
14 carden2b ( 𝐴𝐵 → ( card ‘ 𝐴 ) = ( card ‘ 𝐵 ) )
15 eqimss ( ( card ‘ 𝐴 ) = ( card ‘ 𝐵 ) → ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) )
16 14 15 syl ( 𝐴𝐵 → ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) )
17 16 a1i ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( 𝐴𝐵 → ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) ) )
18 13 17 jaod ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( 𝐴𝐵𝐴𝐵 ) → ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) ) )
19 2 18 syl5bi ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( 𝐴𝐵 → ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) ) )
20 1 19 impbid ( ( 𝐴 ∈ dom card ∧ 𝐵 ∈ dom card ) → ( ( card ‘ 𝐴 ) ⊆ ( card ‘ 𝐵 ) ↔ 𝐴𝐵 ) )