Metamath Proof Explorer


Theorem sspwimp

Description: If a class is a subclass of another class, then its power class is a subclass of that other class's power class. Left-to-right implication of Exercise 18 of TakeutiZaring p. 18. For the biconditional, see sspwb . The proof sspwimp , using conventional notation, was translated from virtual deduction form, sspwimpVD , using a translation program. (Contributed by Alan Sare, 23-Apr-2015) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion sspwimp ⊢ A ⊆ B → 𝒫 A ⊆ 𝒫 B

Proof

Step Hyp Ref Expression
1 vex ⊢ x ∈ V
2 1 a1i ⊢ ⊤ → x ∈ V
3 id ⊢ A ⊆ B → A ⊆ B
4 id ⊢ x ∈ 𝒫 A → x ∈ 𝒫 A
5 elpwi ⊢ x ∈ 𝒫 A → x ⊆ A
6 4 5 syl ⊢ x ∈ 𝒫 A → x ⊆ A
7 sstr ⊢ x ⊆ A ∧ A ⊆ B → x ⊆ B
8 7 ancoms ⊢ A ⊆ B ∧ x ⊆ A → x ⊆ B
9 3 6 8 syl2an ⊢ A ⊆ B ∧ x ∈ 𝒫 A → x ⊆ B
10 2 9 elpwgded ⊢ ⊤ ∧ A ⊆ B ∧ x ∈ 𝒫 A → x ∈ 𝒫 B
11 2 9 10 uun0.1 ⊢ A ⊆ B ∧ x ∈ 𝒫 A → x ∈ 𝒫 B
12 11 ex ⊢ A ⊆ B → x ∈ 𝒫 A → x ∈ 𝒫 B
13 12 alrimiv ⊢ A ⊆ B → ∀ x x ∈ 𝒫 A → x ∈ 𝒫 B
14 df-ss ⊢ 𝒫 A ⊆ 𝒫 B ↔ ∀ x x ∈ 𝒫 A → x ∈ 𝒫 B
15 14 biimpri ⊢ ∀ x x ∈ 𝒫 A → x ∈ 𝒫 B → 𝒫 A ⊆ 𝒫 B
16 13 15 syl ⊢ A ⊆ B → 𝒫 A ⊆ 𝒫 B
17 16 iin1 ⊢ A ⊆ B → 𝒫 A ⊆ 𝒫 B