Theorem eqrrabd

Description: Deduce equality with a restricted abstraction. (Contributed by Thierry Arnoux, 11-Apr-2024)

	Ref	Expression
Hypotheses	eqrrabd.1	⊢ ( 𝜑 → 𝐵 ⊆ 𝐴 )
	eqrrabd.2	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝑥 ∈ 𝐵 ↔ 𝜓 ) )
Assertion	eqrrabd	⊢ ( 𝜑 → 𝐵 = { 𝑥 ∈ 𝐴 ∣ 𝜓 } )

Proof

Step	Hyp	Ref	Expression
1		eqrrabd.1	⊢ ( 𝜑 → 𝐵 ⊆ 𝐴 )
2		eqrrabd.2	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝑥 ∈ 𝐵 ↔ 𝜓 ) )
3		nfv	⊢ Ⅎ 𝑥 𝜑
4		nfcv	⊢ Ⅎ 𝑥 𝐵
5		nfrab1	⊢ Ⅎ 𝑥 { 𝑥 ∈ 𝐴 ∣ 𝜓 }
6	1	sseld	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴 ) )
7	6	pm4.71rd	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐵 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ) )
8	2	pm5.32da	⊢ ( 𝜑 → ( ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) )
9	7 8	bitrd	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐵 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) )
10		rabid	⊢ ( 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) )
11	9 10	bitr4di	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ) )
12	3 4 5 11	eqrd	⊢ ( 𝜑 → 𝐵 = { 𝑥 ∈ 𝐴 ∣ 𝜓 } )