Theorem abss

Description: Class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006)

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

Proof

Step	Hyp	Ref	Expression
1		abid2	⊢ { 𝑥 ∣ 𝑥 ∈ 𝐴 } = 𝐴
2	1	sseq2i	⊢ ( { 𝑥 ∣ 𝜑 } ⊆ { 𝑥 ∣ 𝑥 ∈ 𝐴 } ↔ { 𝑥 ∣ 𝜑 } ⊆ 𝐴 )
3		ss2ab	⊢ ( { 𝑥 ∣ 𝜑 } ⊆ { 𝑥 ∣ 𝑥 ∈ 𝐴 } ↔ ∀ 𝑥 ( 𝜑 → 𝑥 ∈ 𝐴 ) )
4	2 3	bitr3i	⊢ ( { 𝑥 ∣ 𝜑 } ⊆ 𝐴 ↔ ∀ 𝑥 ( 𝜑 → 𝑥 ∈ 𝐴 ) )