Theorem inab

Description: Intersection of two class abstractions. (Contributed by NM, 29-Sep-2002) (Proof shortened by Andrew Salmon, 26-Jun-2011)

	Ref	Expression
Assertion	inab	⊢ ( { 𝑥 ∣ 𝜑 } ∩ { 𝑥 ∣ 𝜓 } ) = { 𝑥 ∣ ( 𝜑 ∧ 𝜓 ) }

Proof

Step	Hyp	Ref	Expression
1		sban	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) )
2		df-clab	⊢ ( 𝑦 ∈ { 𝑥 ∣ ( 𝜑 ∧ 𝜓 ) } ↔ [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) )
3		df-clab	⊢ ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ↔ [ 𝑦 / 𝑥 ] 𝜑 )
4		df-clab	⊢ ( 𝑦 ∈ { 𝑥 ∣ 𝜓 } ↔ [ 𝑦 / 𝑥 ] 𝜓 )
5	3 4	anbi12i	⊢ ( ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ∧ 𝑦 ∈ { 𝑥 ∣ 𝜓 } ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) )
6	1 2 5	3bitr4ri	⊢ ( ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ∧ 𝑦 ∈ { 𝑥 ∣ 𝜓 } ) ↔ 𝑦 ∈ { 𝑥 ∣ ( 𝜑 ∧ 𝜓 ) } )
7	6	ineqri	⊢ ( { 𝑥 ∣ 𝜑 } ∩ { 𝑥 ∣ 𝜓 } ) = { 𝑥 ∣ ( 𝜑 ∧ 𝜓 ) }