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 𝐵 ∧ 𝐴 ∈ 𝐵 ) → ( 1st ‘ 𝐴 ) 𝐵 ( 2nd ‘ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 1st2nd ⊢ ( ( Rel 𝐵 ∧ 𝐴 ∈ 𝐵 ) → 𝐴 = ⟨ ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) ⟩ )
2 simpr ⊢ ( ( Rel 𝐵 ∧ 𝐴 ∈ 𝐵 ) → 𝐴 ∈ 𝐵 )
3 1 2 eqeltrrd ⊢ ( ( Rel 𝐵 ∧ 𝐴 ∈ 𝐵 ) → ⟨ ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) ⟩ ∈ 𝐵 )
4 df-br ⊢ ( ( 1st ‘ 𝐴 ) 𝐵 ( 2nd ‘ 𝐴 ) ↔ ⟨ ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) ⟩ ∈ 𝐵 )
5 3 4 sylibr ⊢ ( ( Rel 𝐵 ∧ 𝐴 ∈ 𝐵 ) → ( 1st ‘ 𝐴 ) 𝐵 ( 2nd ‘ 𝐴 ) )