Theorem ssabf

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

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

Proof

Step	Hyp	Ref	Expression
1		ssabf.1	⊢ Ⅎ 𝑥 𝐴
2	1	abid2f	⊢ { 𝑥 ∣ 𝑥 ∈ 𝐴 } = 𝐴
3	2	sseq1i	⊢ ( { 𝑥 ∣ 𝑥 ∈ 𝐴 } ⊆ { 𝑥 ∣ 𝜑 } ↔ 𝐴 ⊆ { 𝑥 ∣ 𝜑 } )
4		ss2ab	⊢ ( { 𝑥 ∣ 𝑥 ∈ 𝐴 } ⊆ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) )
5	3 4	bitr3i	⊢ ( 𝐴 ⊆ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) )