Theorem inrab

Description: Intersection of two restricted class abstractions. (Contributed by NM, 1-Sep-2006)

	Ref	Expression
Assertion	inrab	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ) = { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) }

Proof

Step	Hyp	Ref	Expression
1		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
2		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) }
3	1 2	ineq12i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ) = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } )
4		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) }
5		inab	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } ) = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) }
6		anandi	⊢ ( ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) )
7	6	abbii	⊢ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) } = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) }
8	5 7	eqtr4i	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } ) = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) }
9	4 8	eqtr4i	⊢ { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) } = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } )
10	3 9	eqtr4i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∩ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ) = { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) }