Metamath Proof Explorer


Theorem 1st2ndbr

Description: Express an element of a relation as a relationship between first and second components. (Contributed by Mario Carneiro, 22-Jun-2016)

Ref Expression
Assertion 1st2ndbr ( ( Rel 𝐵𝐴𝐵 ) → ( 1st𝐴 ) 𝐵 ( 2nd𝐴 ) )

Proof

Step Hyp Ref Expression
1 1st2nd ( ( Rel 𝐵𝐴𝐵 ) → 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ )
2 simpr ( ( Rel 𝐵𝐴𝐵 ) → 𝐴𝐵 )
3 1 2 eqeltrrd ( ( Rel 𝐵𝐴𝐵 ) → ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ ∈ 𝐵 )
4 df-br ( ( 1st𝐴 ) 𝐵 ( 2nd𝐴 ) ↔ ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ ∈ 𝐵 )
5 3 4 sylibr ( ( Rel 𝐵𝐴𝐵 ) → ( 1st𝐴 ) 𝐵 ( 2nd𝐴 ) )