Metamath Proof Explorer


Theorem rabssd

Description: Restricted class abstraction in a subclass relationship. (Contributed by Glauco Siliprandi, 2-Jan-2022)

Ref Expression
Hypotheses rabssd.1 ⊢ Ⅎ 𝑥 𝜑
rabssd.2 ⊢ Ⅎ 𝑥 𝐵
rabssd.3 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ∧ 𝜒 ) → 𝑥 ∈ 𝐵 )
Assertion rabssd ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜒 } ⊆ 𝐵 )

Proof

Step Hyp Ref Expression
1 rabssd.1 ⊢ Ⅎ 𝑥 𝜑
2 rabssd.2 ⊢ Ⅎ 𝑥 𝐵
3 rabssd.3 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ∧ 𝜒 ) → 𝑥 ∈ 𝐵 )
4 3 3exp ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 → ( 𝜒 → 𝑥 ∈ 𝐵 ) ) )
5 1 4 ralrimi ⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝐴 ( 𝜒 → 𝑥 ∈ 𝐵 ) )
6 2 rabssf ⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜒 } ⊆ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜒 → 𝑥 ∈ 𝐵 ) )
7 5 6 sylibr ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜒 } ⊆ 𝐵 )