Metamath Proof Explorer


Theorem bj-rcleq

Description: Relative version of dfcleq . (Contributed by BJ, 27-Dec-2023)

Ref Expression
Assertion bj-rcleq ( ( 𝑉 ∩ 𝐴 ) = ( 𝑉 ∩ 𝐵 ) ↔ ∀ 𝑥 ∈ 𝑉 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 nfcv ⊢ Ⅎ 𝑥 𝐴
2 nfcv ⊢ Ⅎ 𝑥 𝐵
3 nfcv ⊢ Ⅎ 𝑥 𝑉
4 1 2 3 bj-rcleqf ⊢ ( ( 𝑉 ∩ 𝐴 ) = ( 𝑉 ∩ 𝐵 ) ↔ ∀ 𝑥 ∈ 𝑉 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )