Theorem rabss2OLD

Description: Obsolete version of rabss2 as of 1-Feb-2026. (Contributed by NM, 30-May-2006) (Proof shortened by Andrew Salmon, 26-Jun-2011) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	rabss2OLD	⊢ ( 𝐴 ⊆ 𝐵 → { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ { 𝑥 ∈ 𝐵 ∣ 𝜑 } )

Proof

Step	Hyp	Ref	Expression
1		pm3.45	⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
2	1	alimi	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
3		df-ss	⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
4		ss2ab	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } ↔ ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
5	2 3 4	3imtr4i	⊢ ( 𝐴 ⊆ 𝐵 → { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } )
6		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
7		df-rab	⊢ { 𝑥 ∈ 𝐵 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) }
8	5 6 7	3sstr4g	⊢ ( 𝐴 ⊆ 𝐵 → { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ { 𝑥 ∈ 𝐵 ∣ 𝜑 } )