Metamath Proof Explorer


Theorem abeqin

Description: Intersection with class abstraction. (Contributed by Peter Mazsa, 21-Jul-2021)

Ref Expression
Hypotheses abeqin.1 𝐴 = ( 𝐵𝐶 )
abeqin.2 𝐵 = { 𝑥𝜑 }
Assertion abeqin 𝐴 = { 𝑥𝐶𝜑 }

Proof

Step Hyp Ref Expression
1 abeqin.1 𝐴 = ( 𝐵𝐶 )
2 abeqin.2 𝐵 = { 𝑥𝜑 }
3 2 ineq1i ( 𝐵𝐶 ) = ( { 𝑥𝜑 } ∩ 𝐶 )
4 dfrab2 { 𝑥𝐶𝜑 } = ( { 𝑥𝜑 } ∩ 𝐶 )
5 3 1 4 3eqtr4i 𝐴 = { 𝑥𝐶𝜑 }