Theorem ssab

Description: Subclass of a class abstraction. (Contributed by NM, 16-Aug-2006)

	Ref	Expression
Assertion	ssab	⊢ ( 𝐴 ⊆ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		abid2	⊢ { 𝑥 ∣ 𝑥 ∈ 𝐴 } = 𝐴
2	1	sseq1i	⊢ ( { 𝑥 ∣ 𝑥 ∈ 𝐴 } ⊆ { 𝑥 ∣ 𝜑 } ↔ 𝐴 ⊆ { 𝑥 ∣ 𝜑 } )
3		ss2ab	⊢ ( { 𝑥 ∣ 𝑥 ∈ 𝐴 } ⊆ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) )
4	2 3	bitr3i	⊢ ( 𝐴 ⊆ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) )