Theorem rabssf

Description: Restricted class abstraction in a subclass relationship. (Contributed by Glauco Siliprandi, 26-Jun-2021)

	Ref	Expression
Hypothesis	rabssf.1	⊢ Ⅎ 𝑥 𝐵
Assertion	rabssf	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 ∈ 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		rabssf.1	⊢ Ⅎ 𝑥 𝐵
2		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
3	2	sseq1i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ 𝐵 ↔ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ 𝐵 )
4	1	abssf	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ 𝐵 ↔ ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → 𝑥 ∈ 𝐵 ) )
5		impexp	⊢ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → ( 𝜑 → 𝑥 ∈ 𝐵 ) ) )
6	5	albii	⊢ ( ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ( 𝜑 → 𝑥 ∈ 𝐵 ) ) )
7		df-ral	⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ( 𝜑 → 𝑥 ∈ 𝐵 ) ) )
8	6 7	bitr4i	⊢ ( ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 ∈ 𝐵 ) )
9	3 4 8	3bitri	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 ∈ 𝐵 ) )