Metamath Proof Explorer


Theorem 1st2ndb

Description: Reconstruction of an ordered pair in terms of its components. (Contributed by NM, 25-Feb-2014)

Ref Expression
Assertion 1st2ndb ( 𝐴 ∈ ( V × V ) ↔ 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ )

Proof

Step Hyp Ref Expression
1 1st2nd2 ( 𝐴 ∈ ( V × V ) → 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ )
2 id ( 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ → 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ )
3 fvex ( 1st𝐴 ) ∈ V
4 fvex ( 2nd𝐴 ) ∈ V
5 3 4 opelvv ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ ∈ ( V × V )
6 2 5 eqeltrdi ( 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ → 𝐴 ∈ ( V × V ) )
7 1 6 impbii ( 𝐴 ∈ ( V × V ) ↔ 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ )