Theorem ssrabf

Description: Subclass of a restricted class abstraction. (Contributed by Glauco Siliprandi, 26-Jun-2021)

	Ref	Expression
Hypotheses	ssrabf.1	⊢ Ⅎ 𝑥 𝐵
	ssrabf.2	⊢ Ⅎ 𝑥 𝐴
Assertion	ssrabf	⊢ ( 𝐵 ⊆ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ ( 𝐵 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		ssrabf.1	⊢ Ⅎ 𝑥 𝐵
2		ssrabf.2	⊢ Ⅎ 𝑥 𝐴
3		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
4	3	sseq2i	⊢ ( 𝐵 ⊆ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ 𝐵 ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } )
5	1	ssabf	⊢ ( 𝐵 ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )
6	1 2	dfss3f	⊢ ( 𝐵 ⊆ 𝐴 ↔ ∀ 𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 )
7	6	anbi1i	⊢ ( ( 𝐵 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) ↔ ( ∀ 𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) )
8		r19.26	⊢ ( ∀ 𝑥 ∈ 𝐵 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ( ∀ 𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) )
9		df-ral	⊢ ( ∀ 𝑥 ∈ 𝐵 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )
10	7 8 9	3bitr2ri	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ↔ ( 𝐵 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) )
11	4 5 10	3bitri	⊢ ( 𝐵 ⊆ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ ( 𝐵 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) )