Theorem rabrab

Description: Abstract builder restricted to another restricted abstract builder. (Contributed by Thierry Arnoux, 30-Aug-2017)

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

Proof

Step	Hyp	Ref	Expression
1		rabid	⊢ ( 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) )
2	1	anbi1i	⊢ ( ( 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∧ 𝜓 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝜓 ) )
3		anass	⊢ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝜓 ) ↔ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) )
4	2 3	bitri	⊢ ( ( 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∧ 𝜓 ) ↔ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) )
5	4	abbii	⊢ { 𝑥 ∣ ( 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∧ 𝜓 ) } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) }
6		df-rab	⊢ { 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∣ 𝜓 } = { 𝑥 ∣ ( 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∧ 𝜓 ) }
7		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜓 ) ) }
8	5 6 7	3eqtr4i	⊢ { 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∣ 𝜓 } = { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∧ 𝜓 ) }