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 ) )