Metamath Proof Explorer


Theorem bj-inrab3

Description: Generalization of dfrab3ss . Shortens dfrab3ss . (Contributed by BJ, 21-Apr-2019) (Revised by OpenAI, 7-Jul-2020)

Ref Expression
Assertion bj-inrab3 ( 𝐴 ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ) = ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 )

Proof

Step Hyp Ref Expression
1 dfrab3 ⊢ { 𝑥 ∈ 𝐵 ∣ 𝜑 } = ( 𝐵 ∩ { 𝑥 ∣ 𝜑 } )
2 1 ineq2i ⊢ ( 𝐴 ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ) = ( 𝐴 ∩ ( 𝐵 ∩ { 𝑥 ∣ 𝜑 } ) )
3 dfrab3 ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = ( 𝐴 ∩ { 𝑥 ∣ 𝜑 } )
4 3 ineq2i ⊢ ( 𝐵 ∩ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ) = ( 𝐵 ∩ ( 𝐴 ∩ { 𝑥 ∣ 𝜑 } ) )
5 incom ⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 ) = ( 𝐵 ∩ { 𝑥 ∈ 𝐴 ∣ 𝜑 } )
6 in12 ⊢ ( 𝐴 ∩ ( 𝐵 ∩ { 𝑥 ∣ 𝜑 } ) ) = ( 𝐵 ∩ ( 𝐴 ∩ { 𝑥 ∣ 𝜑 } ) )
7 4 5 6 3eqtr4i ⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 ) = ( 𝐴 ∩ ( 𝐵 ∩ { 𝑥 ∣ 𝜑 } ) )
8 2 7 eqtr4i ⊢ ( 𝐴 ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ) = ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 )