Metamath Proof Explorer


Theorem rabss2

Description: Subclass law for restricted abstraction. (Contributed by NM, 18-Dec-2004) (Proof shortened by Andrew Salmon, 26-Jun-2011) Avoid axioms. (Revised by TM, 1-Feb-2026)

Ref Expression
Assertion rabss2 ( 𝐴 ⊆ 𝐵 → { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ { 𝑥 ∈ 𝐵 ∣ 𝜑 } )

Proof

Step Hyp Ref Expression
1 ssel ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
2 1 anim1d ⊢ ( 𝐴 ⊆ 𝐵 → ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
3 2 ss2abdv ⊢ ( 𝐴 ⊆ 𝐵 → { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } )
4 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
5 df-rab ⊢ { 𝑥 ∈ 𝐵 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) }
6 3 4 5 3sstr4g ⊢ ( 𝐴 ⊆ 𝐵 → { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ { 𝑥 ∈ 𝐵 ∣ 𝜑 } )