Theorem rabss

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

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

Proof

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