Metamath Proof Explorer


Theorem intss

Description: Intersection of subclasses. (Contributed by NM, 14-Oct-1999) (Proof shortened by OpenAI, 25-Mar-2020)

Ref Expression
Assertion intss ( 𝐴𝐵 𝐵 𝐴 )

Proof

Step Hyp Ref Expression
1 ssralv ( 𝐴𝐵 → ( ∀ 𝑥𝐵 𝑦𝑥 → ∀ 𝑥𝐴 𝑦𝑥 ) )
2 1 ss2abdv ( 𝐴𝐵 → { 𝑦 ∣ ∀ 𝑥𝐵 𝑦𝑥 } ⊆ { 𝑦 ∣ ∀ 𝑥𝐴 𝑦𝑥 } )
3 dfint2 𝐵 = { 𝑦 ∣ ∀ 𝑥𝐵 𝑦𝑥 }
4 dfint2 𝐴 = { 𝑦 ∣ ∀ 𝑥𝐴 𝑦𝑥 }
5 2 3 4 3sstr4g ( 𝐴𝐵 𝐵 𝐴 )