Theorem inrab2

Description: Intersection with a restricted class abstraction. (Contributed by NM, 19-Nov-2007)

	Ref	Expression
Assertion	inrab2	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 ) = { 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∣ 𝜑 }

Proof

Step	Hyp	Ref	Expression
1		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
2		abid1	⊢ 𝐵 = { 𝑥 ∣ 𝑥 ∈ 𝐵 }
3	1 2	ineq12i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 ) = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ 𝑥 ∈ 𝐵 } )
4		df-rab	⊢ { 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) }
5		inab	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ 𝑥 ∈ 𝐵 } ) = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝑥 ∈ 𝐵 ) }
6		elin	⊢ ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) )
7	6	anbi1i	⊢ ( ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ∧ 𝜑 ) )
8		an32	⊢ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝑥 ∈ 𝐵 ) )
9	7 8	bitri	⊢ ( ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝑥 ∈ 𝐵 ) )
10	9	abbii	⊢ { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) } = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝑥 ∈ 𝐵 ) }
11	5 10	eqtr4i	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ 𝑥 ∈ 𝐵 } ) = { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∧ 𝜑 ) }
12	4 11	eqtr4i	⊢ { 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∣ 𝜑 } = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ 𝑥 ∈ 𝐵 } )
13	3 12	eqtr4i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ 𝐵 ) = { 𝑥 ∈ ( 𝐴 ∩ 𝐵 ) ∣ 𝜑 }