Metamath Proof Explorer


Theorem en2eqpr

Description: Building a set with two elements. (Contributed by FL, 11-Aug-2008) (Revised by Mario Carneiro, 10-Sep-2015)

Ref Expression
Assertion en2eqpr ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) → ( 𝐴𝐵𝐶 = { 𝐴 , 𝐵 } ) )

Proof

Step Hyp Ref Expression
1 2onn 2o ∈ ω
2 nnfi ( 2o ∈ ω → 2o ∈ Fin )
3 1 2 ax-mp 2o ∈ Fin
4 simpl1 ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → 𝐶 ≈ 2o )
5 enfii ( ( 2o ∈ Fin ∧ 𝐶 ≈ 2o ) → 𝐶 ∈ Fin )
6 3 4 5 sylancr ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → 𝐶 ∈ Fin )
7 simpl2 ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → 𝐴𝐶 )
8 simpl3 ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → 𝐵𝐶 )
9 7 8 prssd ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → { 𝐴 , 𝐵 } ⊆ 𝐶 )
10 pr2nelem ( ( 𝐴𝐶𝐵𝐶𝐴𝐵 ) → { 𝐴 , 𝐵 } ≈ 2o )
11 10 3expa ( ( ( 𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → { 𝐴 , 𝐵 } ≈ 2o )
12 11 3adantl1 ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → { 𝐴 , 𝐵 } ≈ 2o )
13 4 ensymd ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → 2o𝐶 )
14 entr ( ( { 𝐴 , 𝐵 } ≈ 2o ∧ 2o𝐶 ) → { 𝐴 , 𝐵 } ≈ 𝐶 )
15 12 13 14 syl2anc ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → { 𝐴 , 𝐵 } ≈ 𝐶 )
16 fisseneq ( ( 𝐶 ∈ Fin ∧ { 𝐴 , 𝐵 } ⊆ 𝐶 ∧ { 𝐴 , 𝐵 } ≈ 𝐶 ) → { 𝐴 , 𝐵 } = 𝐶 )
17 6 9 15 16 syl3anc ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → { 𝐴 , 𝐵 } = 𝐶 )
18 17 eqcomd ( ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) ∧ 𝐴𝐵 ) → 𝐶 = { 𝐴 , 𝐵 } )
19 18 ex ( ( 𝐶 ≈ 2o𝐴𝐶𝐵𝐶 ) → ( 𝐴𝐵𝐶 = { 𝐴 , 𝐵 } ) )