Theorem abssf

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

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

Proof

Step	Hyp	Ref	Expression
1		abssf.1	⊢ Ⅎ 𝑥 𝐴
2	1	abid2f	⊢ { 𝑥 ∣ 𝑥 ∈ 𝐴 } = 𝐴
3	2	sseq2i	⊢ ( { 𝑥 ∣ 𝜑 } ⊆ { 𝑥 ∣ 𝑥 ∈ 𝐴 } ↔ { 𝑥 ∣ 𝜑 } ⊆ 𝐴 )
4		ss2ab	⊢ ( { 𝑥 ∣ 𝜑 } ⊆ { 𝑥 ∣ 𝑥 ∈ 𝐴 } ↔ ∀ 𝑥 ( 𝜑 → 𝑥 ∈ 𝐴 ) )
5	3 4	bitr3i	⊢ ( { 𝑥 ∣ 𝜑 } ⊆ 𝐴 ↔ ∀ 𝑥 ( 𝜑 → 𝑥 ∈ 𝐴 ) )