Metamath Proof Explorer


Theorem opthreg

Description: Theorem for alternate representation of ordered pairs, requiring the Axiom of Regularity ax-reg (via the preleq step). See df-op for a description of other ordered pair representations. Exercise 34 of Enderton p. 207. (Contributed by NM, 16-Oct-1996) (Proof shortened by AV, 15-Jun-2022)

Ref Expression
Hypotheses opthreg.1 𝐴 ∈ V
opthreg.2 𝐵 ∈ V
opthreg.3 𝐶 ∈ V
opthreg.4 𝐷 ∈ V
Assertion opthreg ( { 𝐴 , { 𝐴 , 𝐵 } } = { 𝐶 , { 𝐶 , 𝐷 } } ↔ ( 𝐴 = 𝐶𝐵 = 𝐷 ) )

Proof

Step Hyp Ref Expression
1 opthreg.1 𝐴 ∈ V
2 opthreg.2 𝐵 ∈ V
3 opthreg.3 𝐶 ∈ V
4 opthreg.4 𝐷 ∈ V
5 1 prid1 𝐴 ∈ { 𝐴 , 𝐵 }
6 3 prid1 𝐶 ∈ { 𝐶 , 𝐷 }
7 prex { 𝐴 , 𝐵 } ∈ V
8 7 preleq ( ( ( 𝐴 ∈ { 𝐴 , 𝐵 } ∧ 𝐶 ∈ { 𝐶 , 𝐷 } ) ∧ { 𝐴 , { 𝐴 , 𝐵 } } = { 𝐶 , { 𝐶 , 𝐷 } } ) → ( 𝐴 = 𝐶 ∧ { 𝐴 , 𝐵 } = { 𝐶 , 𝐷 } ) )
9 5 6 8 mpanl12 ( { 𝐴 , { 𝐴 , 𝐵 } } = { 𝐶 , { 𝐶 , 𝐷 } } → ( 𝐴 = 𝐶 ∧ { 𝐴 , 𝐵 } = { 𝐶 , 𝐷 } ) )
10 preq1 ( 𝐴 = 𝐶 → { 𝐴 , 𝐵 } = { 𝐶 , 𝐵 } )
11 10 eqeq1d ( 𝐴 = 𝐶 → ( { 𝐴 , 𝐵 } = { 𝐶 , 𝐷 } ↔ { 𝐶 , 𝐵 } = { 𝐶 , 𝐷 } ) )
12 2 4 preqr2 ( { 𝐶 , 𝐵 } = { 𝐶 , 𝐷 } → 𝐵 = 𝐷 )
13 11 12 syl6bi ( 𝐴 = 𝐶 → ( { 𝐴 , 𝐵 } = { 𝐶 , 𝐷 } → 𝐵 = 𝐷 ) )
14 13 imdistani ( ( 𝐴 = 𝐶 ∧ { 𝐴 , 𝐵 } = { 𝐶 , 𝐷 } ) → ( 𝐴 = 𝐶𝐵 = 𝐷 ) )
15 9 14 syl ( { 𝐴 , { 𝐴 , 𝐵 } } = { 𝐶 , { 𝐶 , 𝐷 } } → ( 𝐴 = 𝐶𝐵 = 𝐷 ) )
16 preq1 ( 𝐴 = 𝐶 → { 𝐴 , { 𝐴 , 𝐵 } } = { 𝐶 , { 𝐴 , 𝐵 } } )
17 16 adantr ( ( 𝐴 = 𝐶𝐵 = 𝐷 ) → { 𝐴 , { 𝐴 , 𝐵 } } = { 𝐶 , { 𝐴 , 𝐵 } } )
18 preq12 ( ( 𝐴 = 𝐶𝐵 = 𝐷 ) → { 𝐴 , 𝐵 } = { 𝐶 , 𝐷 } )
19 18 preq2d ( ( 𝐴 = 𝐶𝐵 = 𝐷 ) → { 𝐶 , { 𝐴 , 𝐵 } } = { 𝐶 , { 𝐶 , 𝐷 } } )
20 17 19 eqtrd ( ( 𝐴 = 𝐶𝐵 = 𝐷 ) → { 𝐴 , { 𝐴 , 𝐵 } } = { 𝐶 , { 𝐶 , 𝐷 } } )
21 15 20 impbii ( { 𝐴 , { 𝐴 , 𝐵 } } = { 𝐶 , { 𝐶 , 𝐷 } } ↔ ( 𝐴 = 𝐶𝐵 = 𝐷 ) )