Metamath Proof Explorer


Theorem eqop2

Description: Two ways to express equality with an ordered pair. (Contributed by NM, 25-Feb-2014)

Ref Expression
Hypotheses eqop2.1 𝐵 ∈ V
eqop2.2 𝐶 ∈ V
Assertion eqop2 ( 𝐴 = ⟨ 𝐵 , 𝐶 ⟩ ↔ ( 𝐴 ∈ ( V × V ) ∧ ( ( 1st𝐴 ) = 𝐵 ∧ ( 2nd𝐴 ) = 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 eqop2.1 𝐵 ∈ V
2 eqop2.2 𝐶 ∈ V
3 1 2 opelvv 𝐵 , 𝐶 ⟩ ∈ ( V × V )
4 eleq1 ( 𝐴 = ⟨ 𝐵 , 𝐶 ⟩ → ( 𝐴 ∈ ( V × V ) ↔ ⟨ 𝐵 , 𝐶 ⟩ ∈ ( V × V ) ) )
5 3 4 mpbiri ( 𝐴 = ⟨ 𝐵 , 𝐶 ⟩ → 𝐴 ∈ ( V × V ) )
6 eqop ( 𝐴 ∈ ( V × V ) → ( 𝐴 = ⟨ 𝐵 , 𝐶 ⟩ ↔ ( ( 1st𝐴 ) = 𝐵 ∧ ( 2nd𝐴 ) = 𝐶 ) ) )
7 5 6 biadanii ( 𝐴 = ⟨ 𝐵 , 𝐶 ⟩ ↔ ( 𝐴 ∈ ( V × V ) ∧ ( ( 1st𝐴 ) = 𝐵 ∧ ( 2nd𝐴 ) = 𝐶 ) ) )