Metamath Proof Explorer


Theorem enpr2d

Description: A pair with distinct elements is equinumerous to ordinal two. (Contributed by Rohan Ridenour, 3-Aug-2023)

Ref Expression
Hypotheses enpr2d.1 ( 𝜑𝐴𝐶 )
enpr2d.2 ( 𝜑𝐵𝐷 )
enpr2d.3 ( 𝜑 → ¬ 𝐴 = 𝐵 )
Assertion enpr2d ( 𝜑 → { 𝐴 , 𝐵 } ≈ 2o )

Proof

Step Hyp Ref Expression
1 enpr2d.1 ( 𝜑𝐴𝐶 )
2 enpr2d.2 ( 𝜑𝐵𝐷 )
3 enpr2d.3 ( 𝜑 → ¬ 𝐴 = 𝐵 )
4 ensn1g ( 𝐴𝐶 → { 𝐴 } ≈ 1o )
5 1 4 syl ( 𝜑 → { 𝐴 } ≈ 1o )
6 1on 1o ∈ On
7 en2sn ( ( 𝐵𝐷 ∧ 1o ∈ On ) → { 𝐵 } ≈ { 1o } )
8 2 6 7 sylancl ( 𝜑 → { 𝐵 } ≈ { 1o } )
9 3 neqned ( 𝜑𝐴𝐵 )
10 disjsn2 ( 𝐴𝐵 → ( { 𝐴 } ∩ { 𝐵 } ) = ∅ )
11 9 10 syl ( 𝜑 → ( { 𝐴 } ∩ { 𝐵 } ) = ∅ )
12 6 onirri ¬ 1o ∈ 1o
13 12 a1i ( 𝜑 → ¬ 1o ∈ 1o )
14 disjsn ( ( 1o ∩ { 1o } ) = ∅ ↔ ¬ 1o ∈ 1o )
15 13 14 sylibr ( 𝜑 → ( 1o ∩ { 1o } ) = ∅ )
16 unen ( ( ( { 𝐴 } ≈ 1o ∧ { 𝐵 } ≈ { 1o } ) ∧ ( ( { 𝐴 } ∩ { 𝐵 } ) = ∅ ∧ ( 1o ∩ { 1o } ) = ∅ ) ) → ( { 𝐴 } ∪ { 𝐵 } ) ≈ ( 1o ∪ { 1o } ) )
17 5 8 11 15 16 syl22anc ( 𝜑 → ( { 𝐴 } ∪ { 𝐵 } ) ≈ ( 1o ∪ { 1o } ) )
18 df-pr { 𝐴 , 𝐵 } = ( { 𝐴 } ∪ { 𝐵 } )
19 df-suc suc 1o = ( 1o ∪ { 1o } )
20 17 18 19 3brtr4g ( 𝜑 → { 𝐴 , 𝐵 } ≈ suc 1o )
21 df-2o 2o = suc 1o
22 20 21 breqtrrdi ( 𝜑 → { 𝐴 , 𝐵 } ≈ 2o )