Metamath Proof Explorer


Theorem pwss

Description: Subclass relationship for power class. (Contributed by NM, 21-Jun-2009)

Ref Expression
Assertion pwss ⊢ 𝒫 A ⊆ B ↔ ∀ x x ⊆ A → x ∈ B

Proof

Step Hyp Ref Expression
1 df-ss ⊢ 𝒫 A ⊆ B ↔ ∀ x x ∈ 𝒫 A → x ∈ B
2 velpw ⊢ x ∈ 𝒫 A ↔ x ⊆ A
3 2 imbi1i ⊢ x ∈ 𝒫 A → x ∈ B ↔ x ⊆ A → x ∈ B
4 3 albii ⊢ ∀ x x ∈ 𝒫 A → x ∈ B ↔ ∀ x x ⊆ A → x ∈ B
5 1 4 bitri ⊢ 𝒫 A ⊆ B ↔ ∀ x x ⊆ A → x ∈ B