Theorem ss2rab

Description: Restricted abstraction classes in a subclass relationship. (Contributed by NM, 30-May-1999)

	Ref	Expression
Assertion	ss2rab	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
2		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) }
3	1 2	sseq12i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ↔ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } )
4		ss2ab	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } ↔ ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) )
5		df-ral	⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝜓 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ( 𝜑 → 𝜓 ) ) )
6		imdistan	⊢ ( ( 𝑥 ∈ 𝐴 → ( 𝜑 → 𝜓 ) ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) )
7	6	albii	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ( 𝜑 → 𝜓 ) ) ↔ ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) )
8	5 7	bitr2i	⊢ ( ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝜓 ) )
9	3 4 8	3bitri	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝜓 ) )