Metamath Proof Explorer


Theorem 1st2nd2

Description: Reconstruction of a member of a Cartesian product in terms of its ordered pair components. (Contributed by NM, 20-Oct-2013)

Ref Expression
Assertion 1st2nd2 ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ )

Proof

Step Hyp Ref Expression
1 elxp6 ( 𝐴 ∈ ( 𝐵 × 𝐶 ) ↔ ( 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ ∧ ( ( 1st𝐴 ) ∈ 𝐵 ∧ ( 2nd𝐴 ) ∈ 𝐶 ) ) )
2 1 simplbi ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ )