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Theorem sspwimpcf 45688
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. sspwimpcf 45688, using conventional notation, was translated from its virtual deduction form, sspwimpcfVD 45689, using a translation program. (Contributed by Alan Sare, 13-Jun-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
sspwimpcf (𝐴𝐵 → 𝒫 𝐴 ⊆ 𝒫 𝐵)

Proof of Theorem sspwimpcf
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 vex 3461 . . . . . 6 𝑥 ∈ V
2 id 23 . . . . . . 7 (𝐴𝐵𝐴𝐵)
3 id 23 . . . . . . . 8 (𝑥 ∈ 𝒫 𝐴𝑥 ∈ 𝒫 𝐴)
4 elpwi 4571 . . . . . . . 8 (𝑥 ∈ 𝒫 𝐴𝑥𝐴)
53, 4syl 18 . . . . . . 7 (𝑥 ∈ 𝒫 𝐴𝑥𝐴)
6 sstr2 3945 . . . . . . . 8 (𝑥𝐴 → (𝐴𝐵𝑥𝐵))
76impcom 413 . . . . . . 7 ((𝐴𝐵𝑥𝐴) → 𝑥𝐵)
82, 5, 7syl2an 608 . . . . . 6 ((𝐴𝐵𝑥 ∈ 𝒫 𝐴) → 𝑥𝐵)
9 elpwg 4567 . . . . . . 7 (𝑥 ∈ V → (𝑥 ∈ 𝒫 𝐵𝑥𝐵))
109biimpar 483 . . . . . 6 ((𝑥 ∈ V ∧ 𝑥𝐵) → 𝑥 ∈ 𝒫 𝐵)
111, 8, 10eel021old 45469 . . . . 5 ((𝐴𝐵𝑥 ∈ 𝒫 𝐴) → 𝑥 ∈ 𝒫 𝐵)
1211ex 418 . . . 4 (𝐴𝐵 → (𝑥 ∈ 𝒫 𝐴𝑥 ∈ 𝒫 𝐵))
1312alrimiv 1960 . . 3 (𝐴𝐵 → ∀𝑥(𝑥 ∈ 𝒫 𝐴𝑥 ∈ 𝒫 𝐵))
14 df-ss 3923 . . . 4 (𝒫 𝐴 ⊆ 𝒫 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝒫 𝐴𝑥 ∈ 𝒫 𝐵))
1514biimpri 231 . . 3 (∀𝑥(𝑥 ∈ 𝒫 𝐴𝑥 ∈ 𝒫 𝐵) → 𝒫 𝐴 ⊆ 𝒫 𝐵)
1613, 15syl 18 . 2 (𝐴𝐵 → 𝒫 𝐴 ⊆ 𝒫 𝐵)
1716iin1 45341 1 (𝐴𝐵 → 𝒫 𝐴 ⊆ 𝒫 𝐵)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wal 1568  wcel 2146  Vcvv 3457  wss 3906  𝒫 cpw 4564
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459  df-ss 3923  df-pw 4566
This theorem is used by: (None)
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